1. In any given circuit first of all recognize the resistances connected in series then by summing the individual resistances draw a new, simplified circuit diagram.
For series combination of resistances equivalent resistance is given by the equation
Req= R1 + R2+R3
The current in each resistor is the same when connected in parallel combination.
2. Then recognize the resistances connected in parallel and find the equivalent resistances of parallel combinations by summing the reciprocals of the resistances and then taking the reciprocal of the result. Draw the new, simplified circuit diagram.
(1/R)=(1/R1)+(1/R2)+(1/R2)
Remember that for resistors connected in parallel combination ‘The potential difference across each resistor is the same’.
3. Repeat the first two steps as required, until no further combinations can be made using resistances. If there is only a single battery in the circuit, this will usually result in a single equivalent resistor in series with the battery.
4. Use Ohm’s Law, V= IR, to determine the current in the equivalent resistor. Then work backwards through the diagrams, applying the useful facts listed in step 1 or step 2 to find the currents in the other resistors. (In more complex circuits, Kirchhoff’s rules can be applied).
Showing posts with label Electricity. Show all posts
Showing posts with label Electricity. Show all posts
Tuesday, 23 November 2010
Wednesday, 22 July 2009
Electric Current and Current Density
- Electric current is the motion of electrically charged particles from one region to another. This motion of electric charges takes place within electric circuit which is a conducting path forming a closed loop.
- In a conductor electric charge will flow from its one end to another if and only if both the ends of the conductor are at different electric potentials.
- Continous flow of electric current in a conductor for a relatively long period of time can be attained using battries which could supply continous flow of charge at low potentials.
Electric current
- Here in this section we would discuss about the electric current in conductors.
- When electric field inside the conductor is zero then there would be no net flow of current in the conductor because in this case electrons in the conductors moves about randomly leaving no net flow of charge in any direction and without the flow of charge there would be no net electric current.
-To maintain a constant current in a conductor we would have to ensure that a constant and steady electric field is established inside the conductor in order to maintain a force on the mobile charges in the conductor.
-Once the electric field is maintained inside the conductor charged particles in the conductors are now under the influence of driving force F = qE.
- In an conductor charged particles undergoes frequent inelastic collision with fixed massive ions in the conductor and undergoes random change in the direction of motion.
- Hence on an average charged particle moves in the direction of driving force with an average velocity known ad drift velocity.
- Electric current is defined as the quantity of charge ΔQ flowing through cross-sectional area A in time interval Δt . Thus,
Iav = ΔQ/Δt
which is the average current flow in time Δt.
- If dQ is the amount of charge flowing in infinitesimally time interval dt through a cross-sectional area of the conductor then instantaneous current I is defined as
I = dQ/dt
- Electric current is a scalar quantity.
- SI unit of current is Ampere (A) where 1A is one coulomb per second.
Drift velocity and Current density - Consider a portion of a conductor of cross-sectional area A. Also consider a small section of conductor of length Δx. Now volume of conductor under consideration is AΔx.
- If there are n number of mobile charge carriers per unit volume then total charge in the section under consideration is
ΔQ = (number of charge carriers in the section) x (charge per carrier)
= (nAΔx)q .................1
where q is the amout of charge on each carrier.
- If charge carriers are moving with speed vd, then distance travelled by the charge carriers is Δx = vdΔt. Putting this in equation 1 we have,
ΔQ = (nAvdΔt)q
- Now current in the conductor is
I = ΔQ/Δt
thus,
I = nqvdA ....................2
where vd is average velocity known as drift velocity as defined earlier.
- Current density j is defined as current per unit cross-sectional area. Thus from 2
j = I/A = nqvd
Current is a scalar quantity but current density can also be defined too include both magnitude and direction. Thus vector current density is
j = nqvd
Direction of current density is same as the direction of electric field.
- Unit of current density is A.m-2.
- Current density tells us about how charges flow at a certain point and also about the direction of the flow at that point but current describes how charges flow throughout an extended object.
Solved example
In this solved example we will lern to apply the principle of current to the problems.
Question:-Amount of charge that passes through a certain conducting wire in 4 sec is 6.5 C. Find (a) the current in the wire and (b) number of electrons that passes through the wire in 8 sec.
Solution : -
(a)
Problem solving strategy
1. From the lesson learned about electric current identify the equation for calculating current when charge and time are given.
2. Put the values of charge and time at respective place and calculate the answer.
In this case Q = 6.5 C and t = 4 sec.
now I = q/t = 6.5C/4 sec = 1.625 A
which is the required answer.
(b)
Problem solving strategy
1. Here we have to find total charge through the wire in a given time.
2. Total charge through the wire would be equal to number of electrons through the circuit multiplied by the charge on each electron.
Here we have to find the number of electrons passing through the wire in 8 sec. If qe is the amount of charge on a single electron and total n number of electrons passes through the wire then
I = nqe/t
or, N = It/qe
putting values of I , t and qe = 1.6 x 10-19 and calculating we get
N = 8.125 x 1019
Same way problems related to current density and drift velocity can be solved.
- In a conductor electric charge will flow from its one end to another if and only if both the ends of the conductor are at different electric potentials.
- Continous flow of electric current in a conductor for a relatively long period of time can be attained using battries which could supply continous flow of charge at low potentials.
Electric current
- Here in this section we would discuss about the electric current in conductors.
- When electric field inside the conductor is zero then there would be no net flow of current in the conductor because in this case electrons in the conductors moves about randomly leaving no net flow of charge in any direction and without the flow of charge there would be no net electric current.
-To maintain a constant current in a conductor we would have to ensure that a constant and steady electric field is established inside the conductor in order to maintain a force on the mobile charges in the conductor.
-Once the electric field is maintained inside the conductor charged particles in the conductors are now under the influence of driving force F = qE.
- In an conductor charged particles undergoes frequent inelastic collision with fixed massive ions in the conductor and undergoes random change in the direction of motion.
- Hence on an average charged particle moves in the direction of driving force with an average velocity known ad drift velocity.
- Electric current is defined as the quantity of charge ΔQ flowing through cross-sectional area A in time interval Δt . Thus,
Iav = ΔQ/Δt
which is the average current flow in time Δt.
- If dQ is the amount of charge flowing in infinitesimally time interval dt through a cross-sectional area of the conductor then instantaneous current I is defined as
I = dQ/dt
- Electric current is a scalar quantity.
- SI unit of current is Ampere (A) where 1A is one coulomb per second.
Drift velocity and Current density - Consider a portion of a conductor of cross-sectional area A. Also consider a small section of conductor of length Δx. Now volume of conductor under consideration is AΔx.
- If there are n number of mobile charge carriers per unit volume then total charge in the section under consideration is
ΔQ = (number of charge carriers in the section) x (charge per carrier)
= (nAΔx)q .................1
where q is the amout of charge on each carrier.
- If charge carriers are moving with speed vd, then distance travelled by the charge carriers is Δx = vdΔt. Putting this in equation 1 we have,
ΔQ = (nAvdΔt)q
- Now current in the conductor is
I = ΔQ/Δt
thus,
I = nqvdA ....................2
where vd is average velocity known as drift velocity as defined earlier.
- Current density j is defined as current per unit cross-sectional area. Thus from 2
j = I/A = nqvd
Current is a scalar quantity but current density can also be defined too include both magnitude and direction. Thus vector current density is
j = nqvd
Direction of current density is same as the direction of electric field.
- Unit of current density is A.m-2.
- Current density tells us about how charges flow at a certain point and also about the direction of the flow at that point but current describes how charges flow throughout an extended object.
Solved example
In this solved example we will lern to apply the principle of current to the problems.
Question:-Amount of charge that passes through a certain conducting wire in 4 sec is 6.5 C. Find (a) the current in the wire and (b) number of electrons that passes through the wire in 8 sec.
Solution : -
(a)
Problem solving strategy
1. From the lesson learned about electric current identify the equation for calculating current when charge and time are given.
2. Put the values of charge and time at respective place and calculate the answer.
In this case Q = 6.5 C and t = 4 sec.
now I = q/t = 6.5C/4 sec = 1.625 A
which is the required answer.
(b)
Problem solving strategy
1. Here we have to find total charge through the wire in a given time.
2. Total charge through the wire would be equal to number of electrons through the circuit multiplied by the charge on each electron.
Here we have to find the number of electrons passing through the wire in 8 sec. If qe is the amount of charge on a single electron and total n number of electrons passes through the wire then
I = nqe/t
or, N = It/qe
putting values of I , t and qe = 1.6 x 10-19 and calculating we get
N = 8.125 x 1019
Same way problems related to current density and drift velocity can be solved.
Friday, 19 September 2008
IITJEE PHYSICS:Current Electricity Part -I
Agenda: Summary of Current Electricity
Some basic point about Electricty
-A current flows along a metal or wire when battery is connected to it.
-Current in a wire is due to free electrons moving along the wire
-The Battery has a Potential difference between its poles due to chemical changes inside the battery.The Potential difference pushes the electrons alongs the metal.The Potential diffrence between the terminal is call the EMF of the battery.
-One pole of the battery is called positive pole and another one is call the negative pole.
-Current direction is opposite to the electron flow direction.
-EMF is measured in Volt
Electric current is defined as the rate at which charge flows inside the conductor
so
I=Q/t
- Its unit is Amphere
Also I=dQ/dt
Drift speed of electrons inside the conductor
vd=I/neA
Where I is the current,n is the number of the electron per unit volume ,e is the electronic charge and A is the area of the crossection of the conductor
Ohms Law
V=IR
where V is the Potential difference accross its ends
I is the current flow
And R is the electrical Resistance of the conductor.Electrical Resistance unit is Ohm
ELectrical Resistivity
It is defined as
ρ=RA/L
Where R is the Resistance
A is the crossectional Area
L is the length
Series Resistors
R=R1+R2+R3
Parallel Resistors
(1/R)=(1/R1)+(1/R2)+(1/R2)
Terminal Voltage of the Cell
V=E-Ir
where E is the EMF of the cell and r is the internal resistance of the Cell
Please let us know if you like this post and what can be done to improve on that
Please comments
Some basic point about Electricty
-A current flows along a metal or wire when battery is connected to it.
-Current in a wire is due to free electrons moving along the wire
-The Battery has a Potential difference between its poles due to chemical changes inside the battery.The Potential difference pushes the electrons alongs the metal.The Potential diffrence between the terminal is call the EMF of the battery.
-One pole of the battery is called positive pole and another one is call the negative pole.
-Current direction is opposite to the electron flow direction.
-EMF is measured in Volt
Electric current is defined as the rate at which charge flows inside the conductor
so
I=Q/t
- Its unit is Amphere
Also I=dQ/dt
Drift speed of electrons inside the conductor
vd=I/neA
Where I is the current,n is the number of the electron per unit volume ,e is the electronic charge and A is the area of the crossection of the conductor
Ohms Law
V=IR
where V is the Potential difference accross its ends
I is the current flow
And R is the electrical Resistance of the conductor.Electrical Resistance unit is Ohm
ELectrical Resistivity
It is defined as
ρ=RA/L
Where R is the Resistance
A is the crossectional Area
L is the length
Series Resistors
R=R1+R2+R3
Parallel Resistors
(1/R)=(1/R1)+(1/R2)+(1/R2)
Terminal Voltage of the Cell
V=E-Ir
where E is the EMF of the cell and r is the internal resistance of the Cell
Please let us know if you like this post and what can be done to improve on that
Please comments
Monday, 11 August 2008
Current electricity test series
(A) Assertion reason type question.
(a) If both assertion and reason are true and reason is correct explaination of assertion.
(b) If both assertion and reason are true and reason is not correct explaination of assertion.
(c) Assertion is true but reason is false.
(d) both assertion and reason are false.
(1)
Assertion:Wire carrying current is not charged.
Reason:It is because , at any instant number of electrons leaving wire is sometimes equal to the number of electrons flowing in from the battery.
(2)
Assertion:Drift velocity of electrons decreases on increasing temperature of the conductor.
Reason:It is because, on increasing temperature of a conductor , the value of resistivity of it's material increases.
(3)
Assertion:Resistors are connected in series combination in order to increase the resistance of the circuit.
Reason:In series combination potential difference across any resistor is proportional to it's resistance.
(4)
Assertion:A potentiometer measures the potential difference more accurately then a voltmeter.
Reason:Because it has a wire of high resistance and draws a heavy current from external circuit.
(B) Multiple choice questions with only one answer
(5) A constant voltage is applied between the two ends of a uniform metalic wire and some heat is developed in it . The heat developed is doubled if,
(a) both length and radius of wire are halved
(b) both length and radius of wire are doubled
(c) radius of wire is doubled
(d) length of wire is doubled
(6) Kirchoff's first law of electric circuits is based on the law of conservation of
(a) only on mass
(b) only on charge
(c) on charge as well as on energy
(d) on charge as well as on mass
(7)The temperature coefficent of resistance of wire is 0.00125°C-1. At 300K, it's resistance is 1 ohm. The resistance of wire would be 2 ohm at
(a) 1154K
(b) 1127K
(c) 1167K
(d) 1176K
(8) If the length of the potentiometer wire is increased, then the accuracy in determination of null point
(a) will increase
(b) will decrease
(c) will remain unaffected
(d) can not be decided , untill emf of auxiliary battery is known
(9) Each of the resistance in the network shown in the figure is equal to R. The resistance between terminal A and B is
(a) R
(b) 5R
(c) 3R
(d) 6R
(10)The current voltage graph for a given metallic wire at two different temperatures T1 and T2 are shown in the fig. which one of the following option is true
(a) T1 = T2
(b) T1 < T2
(c) T1 > T2
(d) none of the above
Solutions
1.c
2.a
3.b
4.c
5.b
6.b
7.b
8.a
9.a
10.b
(a) If both assertion and reason are true and reason is correct explaination of assertion.
(b) If both assertion and reason are true and reason is not correct explaination of assertion.
(c) Assertion is true but reason is false.
(d) both assertion and reason are false.
(1)
Assertion:Wire carrying current is not charged.
Reason:It is because , at any instant number of electrons leaving wire is sometimes equal to the number of electrons flowing in from the battery.
(2)
Assertion:Drift velocity of electrons decreases on increasing temperature of the conductor.
Reason:It is because, on increasing temperature of a conductor , the value of resistivity of it's material increases.
(3)
Assertion:Resistors are connected in series combination in order to increase the resistance of the circuit.
Reason:In series combination potential difference across any resistor is proportional to it's resistance.
(4)
Assertion:A potentiometer measures the potential difference more accurately then a voltmeter.
Reason:Because it has a wire of high resistance and draws a heavy current from external circuit.
(B) Multiple choice questions with only one answer
(5) A constant voltage is applied between the two ends of a uniform metalic wire and some heat is developed in it . The heat developed is doubled if,
(a) both length and radius of wire are halved
(b) both length and radius of wire are doubled
(c) radius of wire is doubled
(d) length of wire is doubled
(6) Kirchoff's first law of electric circuits is based on the law of conservation of
(a) only on mass
(b) only on charge
(c) on charge as well as on energy
(d) on charge as well as on mass
(7)The temperature coefficent of resistance of wire is 0.00125°C-1. At 300K, it's resistance is 1 ohm. The resistance of wire would be 2 ohm at
(a) 1154K
(b) 1127K
(c) 1167K
(d) 1176K
(8) If the length of the potentiometer wire is increased, then the accuracy in determination of null point
(a) will increase
(b) will decrease
(c) will remain unaffected
(d) can not be decided , untill emf of auxiliary battery is known
(9) Each of the resistance in the network shown in the figure is equal to R. The resistance between terminal A and B is
(a) R
(b) 5R
(c) 3R
(d) 6R
(10)The current voltage graph for a given metallic wire at two different temperatures T1 and T2 are shown in the fig. which one of the following option is true
(a) T1 = T2
(b) T1 < T2
(c) T1 > T2
(d) none of the above
Solutions
1.c
2.a
3.b
4.c
5.b
6.b
7.b
8.a
9.a
10.b
Saturday, 14 June 2008
Capacitors
-Capacitors are important components used in electronics and telecommunication devices for example radio , television recivers , transmitter circuits etc.
-Capacitor is a device used for storing electronic charge.
-All capacitors consists of two metal plates(or conductors) separated by an insulator(air, vacuum or any other dielectric medium).
-Conventional symbol of capacitor is where T is the terminal (positive or negative) of battery joined to the plates.
-Capacitor gets charged when a battery is connected to it or when there is a potential difference between two metal plates of the capacitor.
-Capacitor gets discharged on joining two of it's plates.
-If V is the potential difference between two plates of the capacitor and q is the amount of charge developed on each plate then q/V is constant for the capacitor since q∝V.
-The ratio of charge on either plate to the potential difference between the plates is called capacitance C of the capacitor. Thus,
C=q/V
or q=CV
-Unit of capacitance is Farads(F) or CV-1.
-1F is very large unit of capacitance. Practically capacitors with capacitance of the order of micro farads (μF) are used in circuits of radio recivers , transmitters etc. Thus,
1μF=10-6 (micro)
1nF=10-9 (nano)
1pF=10-12 (pico)
-For any capacitor it's capacitance is constant and depends on shape , size , separation f the two conductors and also on insulating medium being used for making capacitor.
- Capacitance of parallel plate capacitor havinf vacuum or air acting as dielectric or insulating medium is
C=(ε0A)/d
where,
C= capacitance of capacitor
A= area of conducting plate
d= distance between plates of the capacitor
ε0=8.854× 10-12 and is known as electric permitivitty in vacuum.
-If k is the relative permittivity of the dielectric medium then
ε=ε0k
thus capacitance of parallel plate air capacitor in presence of dielectric medium of electric permitivity ε is
C=εA/d
-Capacitance of spherical capacitor having radii a, b (b>a) with
(a) air as dielectric between them
C=(4πε0ab)/(b-a)
(b) dielectric with relative permitivity ε
C=(4πεab)/(b-a)
-Capacitor is a device used for storing electronic charge.
-All capacitors consists of two metal plates(or conductors) separated by an insulator(air, vacuum or any other dielectric medium).
-Conventional symbol of capacitor is where T is the terminal (positive or negative) of battery joined to the plates.
-Capacitor gets charged when a battery is connected to it or when there is a potential difference between two metal plates of the capacitor.
-Capacitor gets discharged on joining two of it's plates.
-If V is the potential difference between two plates of the capacitor and q is the amount of charge developed on each plate then q/V is constant for the capacitor since q∝V.
-The ratio of charge on either plate to the potential difference between the plates is called capacitance C of the capacitor. Thus,
C=q/V
or q=CV
-Unit of capacitance is Farads(F) or CV-1.
-1F is very large unit of capacitance. Practically capacitors with capacitance of the order of micro farads (μF) are used in circuits of radio recivers , transmitters etc. Thus,
1μF=10-6 (micro)
1nF=10-9 (nano)
1pF=10-12 (pico)
-For any capacitor it's capacitance is constant and depends on shape , size , separation f the two conductors and also on insulating medium being used for making capacitor.
- Capacitance of parallel plate capacitor havinf vacuum or air acting as dielectric or insulating medium is
C=(ε0A)/d
where,
C= capacitance of capacitor
A= area of conducting plate
d= distance between plates of the capacitor
ε0=8.854× 10-12 and is known as electric permitivitty in vacuum.
-If k is the relative permittivity of the dielectric medium then
ε=ε0k
thus capacitance of parallel plate air capacitor in presence of dielectric medium of electric permitivity ε is
C=εA/d
-Capacitance of spherical capacitor having radii a, b (b>a) with
(a) air as dielectric between them
C=(4πε0ab)/(b-a)
(b) dielectric with relative permitivity ε
C=(4πεab)/(b-a)
Thursday, 12 June 2008
IITJEE Test series Electric Potential
Multiple choice question with only one answer
1.The elctrical potential energy of an islolated metal sphere of radius R and total Charge Q
a. Q2/4πεR
b. Q2/8πεR
c. Q2/2πεR
d. Q2/16πεR
2. A electric dipole is placed at +q(-a,0) and -q(a,o) in the xy plane.Find the workdone by the electric lines on charge which is moved from point (0,a) to (0,-a).
a. zero
b. p/4πεr2
c. -p/4πεr2
d. none of the above
3.Find the electric field at the centre of the uniformly charged semicircular arc of of Radius R and linear charge density λ
a.λ/2πεa
b.λ/4πεa
c. λ/πεa
d. none of these
4.An electric dipole placed in a uniform electric field experiences
a. A torque but no force
b A force but no torque
c. Both force and torque
d neither a force and torque
5.A charge q is placed at the center of the line joining two equal charges Q.The system of three charges will be in equilibrium if q equal is
a. -Q/2
b. Q/2
c. Q/4
d. -Q/4
Multiple choice questions with one or more answer
6.Choose the correct statement
a.if electric field is zero at the point then electric potential must be also zero at that point
b.Two diffrent equipotential surface can intersect
c.if electric potential is constant in a given region then electric field must be zero in that region
d. Electrons move from higher potential to lower potential
7.A spherical conductor shell has charge Q on it and Radius of the spherical shell is R
a. Electric potential decrease with 0< r < infinity
b Electric field decrease with with 0< r < infinity
c Electric potential is non zero constant for 0< r < =R and decrease for R< r < infinity
d Electric field is zero for 0< r < =R and decrease for R< r < infinity
8.Electric potential V(x,y) of a electrostatic field E=a(yi +xj) where a is constant
a.V(x,y)-V(0,0)=-axy
b.V(x,y)-V(1,1)=-axy+a
c.V(x,y)-V(1,1)=-axy-a
d. None of the these
Assertion and Reason
a) Statement I is true ,statement II is true ,statement II is correct explanation for statement I
b) Statement I is true ,statement II is true ,statement II is not a correct explanation for statement I
c) Statement I is true,Statement II is false
d) Statement I is False,Statement II is True
9.
STATEMENT I:Electrix flux through any closed surface around point charge is independent of the size and shape
STATEMENT II.φ=∫ E.da
10.
STATEMENT I:Electric potential inside the spherical conductor shell is nonzero constant
STATEMENT II:Electric field inside the shell is zero
11.
STATEMENT I:Electric Field on the surface of a conductor is less at the sharp corners
STATEMENT II: Surface charge density on conductor surface is inversely proportional to the radius of curvature
1.The elctrical potential energy of an islolated metal sphere of radius R and total Charge Q
a. Q2/4πεR
b. Q2/8πεR
c. Q2/2πεR
d. Q2/16πεR
2. A electric dipole is placed at +q(-a,0) and -q(a,o) in the xy plane.Find the workdone by the electric lines on charge which is moved from point (0,a) to (0,-a).
a. zero
b. p/4πεr2
c. -p/4πεr2
d. none of the above
3.Find the electric field at the centre of the uniformly charged semicircular arc of of Radius R and linear charge density λ
a.λ/2πεa
b.λ/4πεa
c. λ/πεa
d. none of these
4.An electric dipole placed in a uniform electric field experiences
a. A torque but no force
b A force but no torque
c. Both force and torque
d neither a force and torque
5.A charge q is placed at the center of the line joining two equal charges Q.The system of three charges will be in equilibrium if q equal is
a. -Q/2
b. Q/2
c. Q/4
d. -Q/4
Multiple choice questions with one or more answer
6.Choose the correct statement
a.if electric field is zero at the point then electric potential must be also zero at that point
b.Two diffrent equipotential surface can intersect
c.if electric potential is constant in a given region then electric field must be zero in that region
d. Electrons move from higher potential to lower potential
7.A spherical conductor shell has charge Q on it and Radius of the spherical shell is R
a. Electric potential decrease with 0< r < infinity
b Electric field decrease with with 0< r < infinity
c Electric potential is non zero constant for 0< r < =R and decrease for R< r < infinity
d Electric field is zero for 0< r < =R and decrease for R< r < infinity
8.Electric potential V(x,y) of a electrostatic field E=a(yi +xj) where a is constant
a.V(x,y)-V(0,0)=-axy
b.V(x,y)-V(1,1)=-axy+a
c.V(x,y)-V(1,1)=-axy-a
d. None of the these
Assertion and Reason
a) Statement I is true ,statement II is true ,statement II is correct explanation for statement I
b) Statement I is true ,statement II is true ,statement II is not a correct explanation for statement I
c) Statement I is true,Statement II is false
d) Statement I is False,Statement II is True
9.
STATEMENT I:Electrix flux through any closed surface around point charge is independent of the size and shape
STATEMENT II.φ=∫ E.da
10.
STATEMENT I:Electric potential inside the spherical conductor shell is nonzero constant
STATEMENT II:Electric field inside the shell is zero
11.
STATEMENT I:Electric Field on the surface of a conductor is less at the sharp corners
STATEMENT II: Surface charge density on conductor surface is inversely proportional to the radius of curvature
Monday, 2 June 2008
Electric Flux and Gauss law
Electric Flux
dφ=E.da
da is the area vector to the surface and it is taken +ve along the outward normal to the surface
dφ=Edacosθ
φ=∫ E.da
For closed surface
φ=∫ E.da
Guass Theorem
Flux in closed surface is equal net charge inside divided by ε
∫ E.da=qin/ε
Some points:
a. E is the electric field present due to all charges in the ssystem not just the charge inside
b.Flux crossing a closed surface does not depend on the shapes and size of gaussian surface
Electric Potential energy of a charge
=qV where V is the potential there
Others important things
1. ∫ E.dl over closed path is zero
2.Electric potential in the spherical charge conductor is Q/4πεR where R is the radius of the shell and the potential is same everywhere in the conductor
3 Conductor surface is a equipotential surface
dφ=E.da
da is the area vector to the surface and it is taken +ve along the outward normal to the surface
dφ=Edacosθ
φ=∫ E.da
For closed surface
φ=∫ E.da
Guass Theorem
Flux in closed surface is equal net charge inside divided by ε
∫ E.da=qin/ε
Some points:
a. E is the electric field present due to all charges in the ssystem not just the charge inside
b.Flux crossing a closed surface does not depend on the shapes and size of gaussian surface
Electric Potential energy of a charge
=qV where V is the potential there
Others important things
1. ∫ E.dl over closed path is zero
2.Electric potential in the spherical charge conductor is Q/4πεR where R is the radius of the shell and the potential is same everywhere in the conductor
3 Conductor surface is a equipotential surface
Sunday, 18 May 2008
IIT JEE Test Series(Electricity-1)
Multiple choice question with only one answer
1.A rod lies along the x-axis with one end at the origin and other at x->∞it caries a uniform charge λ C/m.Find the electric field at the point x=-a on the x-axis
a. -(λ/4πε0a)i
b.-(λ/4πε0a2)i
c. (λ/4πε0a)i
d. (λ/4πε0a2)i
2.Twelve charges of charge q are situated at the corners of the 12 sided polygon of side a.What is the net force on the charge Q at the center
a. Zero
b. 3qQ/πε0a2
c. qQ/πε0a2
d None of the above
3.Two positive point charge are placed at the distance a apart have sum Q.What values of the charges,coulumb force between them is maximum
a.q1=q1=Q/2
b. q1=3Q/4 ,q2=Q/4
c. q1=5Q/6 ,q2=Q/6
d Non of the above
4.A metallic shell having inner radius R1 and outer radii R2 has a point charge Q kept inside the cavity.Electric field in the region R1 < r < R2 where r is the distance from the center is given by
a depends on the value of r
b zero
c, Constant and nonzero everywhere
d. None of the above
5.Consider two statements
A)The force with which two charges interact is not changed by the presence of the other charges
B)Electric force expierenced by the charge particle due to number of fixed point charges is vector resultant of the forces expierence due to individual charges
a) A and B both are correct
b) A is correct only
c) B is correct only
d. A and B both are wrong
Multiple choice question with more than one answer
6)A metallic sold sphere of radius R is given the charge Q.Which of the following statement is true then
a) Electric field at points 0< r < R is zero
b) Charge Q is on the outer surface of the sphere
c) Electric field at r>R is given by Q/4πε0r2
d) Electric field is perpendicular to the surface of the sphere
7)A simple pendulum consists of a small sphere of mass and posutive charge q is suspended by the string of length L.The pendulum is placed in the electric field of strength E directed vertically downwards.Which of the following is true
a)Time period of oscillation=T=2π√L/(g+qE/m)
b)Time period of oscillation=T=2π√L/(g-qE/m)
c) Tension in the string when the pendulum is at rest =mq+qE
d)Tension in the string when the pendulum is at rest =mq-qE
8)A rod lies on the x-axis with end end at x=-L and other end at x=L with uniform charge λ C/m.Which of the following is true
a.Electric field at any point (0,y) on the y-axis is given by
E=(2kλL/y√(y2+L2))j
b. For point on the Y-axis greater than y>>>>L
E=(2kλL/y2)j
c. Electric field if L->∞
E=2kλ/y
d. None of the above
9. A particle of mass m and charge q is thrown horizontally with a velocity v from top of the building of height H.An electric field exists in the plane and it is horizontally away from the building
which of the following is true
a) Range of the particle is greater than v√(2H/g)
b) Time of flight is √(2H/g)
c) Path is parabolic
d None of the above
Assertion and Reason
a) Statement I is true ,statement II is true ,statement II is correct explanation for statement I
b) Statement I is true ,statement II is true ,statement II is not a correct explanation for statement I
c) Statement I is true,Statement II is false
d) Statement I is False,Statement II is True
10.
STATEMENT I:There is no electric field in conducter in electrostatics
STATEMENT II:There are plenty of free electron in the conductor which moves in such a way to cancel all the electric field
11.
STATEMENT I:Coulumb law of forces are a action reaction pair
STATEMENT II:Coulumb law of forces are non conservative in nature
12.
STATEMENT I:Two electric lines of forces never intersect
STATEMENT II:Electric lines of forces originate from -ve charge and terminate on +ve charge
13.
STATEMENT I:Excess charge given to conductor always reside on the outer surface
STATEMENT II:E is zero inside the bulk material of the conductor
Matrix Match type
14) A metallic shell is having a charge q at the center.Column I gives the location of the charge,Column II defines the charge distribution on the surface due to the charge.You have to match the statement in Column I to column II
Column I
A) q is at the center
B) q is displaced from the center
C) An excess charge Q is given to the metallic shell
D) An Charge Q is placed near the shell
Column II
P)Charge distribution is uniform at the inner surface of the metallic shell
Q)Charge distribution is non uniform at the inner surface of the metallic shell
R)Charge distribution is uniform at the outer surface of the metallic shell
S)Charge distribution is non uniform at the outer surface of the metallic shell
Linked Comprehension Type
A electric charge q of mass m is projected horizontally with a velocity v in the Uniform elctric field which occupies limited space in the region.Electric field E is vertically upwards.And the limited region of the electric field is of length a.Assuming X axis as horizontal and Y-axis as vertical.And let us assumes charge start from origin in the electric field.
Answer following question based on this
15)Horizontal components and Vertical components of the velocity when it passes through the end of the field i.e at x=a
a)v,qEa/mv
b)v,qE/mv
c)qEa/mv,a
d)qE/mv,v
16) Equation of the motion of the particle
a) staright line
b) parabola
c) circle
d) none of the above
17) if the charge get deflected by angle α when it comes out of electric field ,then which one is true
a) tanα=qEa/mv2
b) tanα=qEa/mv
c) sinα=qEa/mv
d) sinα=qEa/mv2
Solutions
1.A rod lies along the x-axis with one end at the origin and other at x->∞it caries a uniform charge λ C/m.Find the electric field at the point x=-a on the x-axis
a. -(λ/4πε0a)i
b.-(λ/4πε0a2)i
c. (λ/4πε0a)i
d. (λ/4πε0a2)i
2.Twelve charges of charge q are situated at the corners of the 12 sided polygon of side a.What is the net force on the charge Q at the center
a. Zero
b. 3qQ/πε0a2
c. qQ/πε0a2
d None of the above
3.Two positive point charge are placed at the distance a apart have sum Q.What values of the charges,coulumb force between them is maximum
a.q1=q1=Q/2
b. q1=3Q/4 ,q2=Q/4
c. q1=5Q/6 ,q2=Q/6
d Non of the above
4.A metallic shell having inner radius R1 and outer radii R2 has a point charge Q kept inside the cavity.Electric field in the region R1 < r < R2 where r is the distance from the center is given by
a depends on the value of r
b zero
c, Constant and nonzero everywhere
d. None of the above
5.Consider two statements
A)The force with which two charges interact is not changed by the presence of the other charges
B)Electric force expierenced by the charge particle due to number of fixed point charges is vector resultant of the forces expierence due to individual charges
a) A and B both are correct
b) A is correct only
c) B is correct only
d. A and B both are wrong
Multiple choice question with more than one answer
6)A metallic sold sphere of radius R is given the charge Q.Which of the following statement is true then
a) Electric field at points 0< r < R is zero
b) Charge Q is on the outer surface of the sphere
c) Electric field at r>R is given by Q/4πε0r2
d) Electric field is perpendicular to the surface of the sphere
7)A simple pendulum consists of a small sphere of mass and posutive charge q is suspended by the string of length L.The pendulum is placed in the electric field of strength E directed vertically downwards.Which of the following is true
a)Time period of oscillation=T=2π√L/(g+qE/m)
b)Time period of oscillation=T=2π√L/(g-qE/m)
c) Tension in the string when the pendulum is at rest =mq+qE
d)Tension in the string when the pendulum is at rest =mq-qE
8)A rod lies on the x-axis with end end at x=-L and other end at x=L with uniform charge λ C/m.Which of the following is true
a.Electric field at any point (0,y) on the y-axis is given by
E=(2kλL/y√(y2+L2))j
b. For point on the Y-axis greater than y>>>>L
E=(2kλL/y2)j
c. Electric field if L->∞
E=2kλ/y
d. None of the above
9. A particle of mass m and charge q is thrown horizontally with a velocity v from top of the building of height H.An electric field exists in the plane and it is horizontally away from the building
which of the following is true
a) Range of the particle is greater than v√(2H/g)
b) Time of flight is √(2H/g)
c) Path is parabolic
d None of the above
Assertion and Reason
a) Statement I is true ,statement II is true ,statement II is correct explanation for statement I
b) Statement I is true ,statement II is true ,statement II is not a correct explanation for statement I
c) Statement I is true,Statement II is false
d) Statement I is False,Statement II is True
10.
STATEMENT I:There is no electric field in conducter in electrostatics
STATEMENT II:There are plenty of free electron in the conductor which moves in such a way to cancel all the electric field
11.
STATEMENT I:Coulumb law of forces are a action reaction pair
STATEMENT II:Coulumb law of forces are non conservative in nature
12.
STATEMENT I:Two electric lines of forces never intersect
STATEMENT II:Electric lines of forces originate from -ve charge and terminate on +ve charge
13.
STATEMENT I:Excess charge given to conductor always reside on the outer surface
STATEMENT II:E is zero inside the bulk material of the conductor
Matrix Match type
14) A metallic shell is having a charge q at the center.Column I gives the location of the charge,Column II defines the charge distribution on the surface due to the charge.You have to match the statement in Column I to column II
Column I
A) q is at the center
B) q is displaced from the center
C) An excess charge Q is given to the metallic shell
D) An Charge Q is placed near the shell
Column II
P)Charge distribution is uniform at the inner surface of the metallic shell
Q)Charge distribution is non uniform at the inner surface of the metallic shell
R)Charge distribution is uniform at the outer surface of the metallic shell
S)Charge distribution is non uniform at the outer surface of the metallic shell
Linked Comprehension Type
A electric charge q of mass m is projected horizontally with a velocity v in the Uniform elctric field which occupies limited space in the region.Electric field E is vertically upwards.And the limited region of the electric field is of length a.Assuming X axis as horizontal and Y-axis as vertical.And let us assumes charge start from origin in the electric field.
Answer following question based on this
15)Horizontal components and Vertical components of the velocity when it passes through the end of the field i.e at x=a
a)v,qEa/mv
b)v,qE/mv
c)qEa/mv,a
d)qE/mv,v
16) Equation of the motion of the particle
a) staright line
b) parabola
c) circle
d) none of the above
17) if the charge get deflected by angle α when it comes out of electric field ,then which one is true
a) tanα=qEa/mv2
b) tanα=qEa/mv
c) sinα=qEa/mv
d) sinα=qEa/mv2
Solutions
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