Electric dipole problem

Question:
A point electric dipole having dipole moment p is placed in an external uniform electric field such that directions of dipole moment and electric field coincides. Find the radius of the sphere which forms one of the equi-potential surface enclosing the dipole.

Solution:

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Electrostatic question 1 (electric potential)

Question:
There are two identical particles each of mass m crying a charge Q. Initially one of them is at rest and another charge moves with velocity v directly towards free particle. Find the distance of closest approach.
Solution:
This is an easy question what you simply have to do is apply the law of conservation of momentum and energy keeping in mind that velocity of both the particles are equal at the distance of closest approach. First try the question yourself and then look for the solution.
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Numerical for class 12 on electric field and potential

Question:
A thin stationary ring of 1 m has a positive charge of 1 x 10^-5C uniformly distributed over it. A particle of mass 0.9 gm and having a negative charge of 1 x 10^-6 C is placed on the axis at a distance 1 cm from the center of the ring. Show that the motion of the negatively charged particle is approximately simple harmonic. Calculate the time period of oscillation.

Solution:
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What are Dielectrics

• A dielectric is an insulating material in which all nuclei are tightly bound to the nuclei of the atom.
• There are no free electrons in a dielectric material so they can not carry current.
• Electric conductivity of a dielectric material is very low and that of an ideal dielectric is zero.
• Some examples of dielectric materials are glass, plastic, mica, oil etc.
• Dielectric constant of a material is the ratio of the capacitance of the capacitor completely filled with the material to the capacitance of the same capacitor in the vacuum i.e. dielectric constant K=C/C₀.
• If the space between a capacitor is filled with a dielectric material capacitance of the capacitor increases.
• The value of dielectric constant K is independent of the shape and size of the capacitor and its value is different for different insulating materials.

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CAPACITOR question with solution (Difficulty level : easy one)

Question-1:A Parallel Plate capacitor has following dimensions
Distance between the plates=10 cm
Area of Plate=2 m²
Charge on each plate=8.85 x 10^-10 C
Calculate following

1. Electric Field outside the plates
2. Electric Field Between the plates
3. Capacitance of the capacitor
4. Energy stored in the capacitor

ε₀=8.854 x 10^-12 C²N^-1m^-2
Solution-1:

As we know that Electric field outside the plates are zero
Electric field Inside the plates is
E=Q/ε₀A =50NC^-1

Capacitance=ε₀A/d =8.854 x 10^-12  x 2/.1=17.6 x 10^-11 F

Energy stored in capacitor=(1/2)Q²/C=.5 x 8.85 x 10^-10 x 8.85 x 10^-10/17.6 x 10^-11
=22.125 x 10^-10 J

Challenging problem in capacitance

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Electrostatics :- Very Short answer type questions

Question 1: Define following terms

(a) Dielectric constant of a medium
(b) Electric dipole moment
(c) Electric flux
Answer 1:
(a)It is the ratio of the capacitance (C_r) of a capacitor with dielectric between the plates to the capacitance (C_r) of the same capacitor with vacuum between the plates i.e. K=C_r/C₀
(b) It is the product of the magnitude of one of the point charges constitting the dipole ant the distance of separation between two point charges.
(c) Electric flux through an area is the product of the magnitude of the area and the component of electric field vector normal to this area element.
Electric flux = E.ds
SI unit of flux is NC¹m²

Question 2: Electric field inside a dielectric decreases when it is placed in an external field. Give reason to support this statement.
Answer 2: An electric field E_P is induced inside the dielectric in a direction opposite to the direction of external electric field E₀. Thus net field becomes
E=E₀-E_P

Question 3: Electric field lines can not be discontinuous. Give reason.
Answer 3: Electric field lines can not be discontinous because if they are discontinous then it will indicate the absence of electric field at the break point.

Question 4: Why do electric field lines can never cross each other?
Answer 4: It is so because if they cross each other then at the point of intersection there will be two tangents which is not possible.

Question 5: What is the net amount of charge on a charged capacitor?

Answer 5: The net charge of a charged capacitor is zero because the charges on its two plates are equal in number and opposite in sign. Even when the capacitor is discharged net charge of the capacitor remains zero because then each plate has zero charge.

Cylendrical capacitor

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Coulumb's law

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Overview of electrostatics and electricity

For a note on overview of electrostatics and electricity click the link given below:
Overview of electrostatics and electricity:
"Electrostatics involves electric charges namely positive and negative charges, the forces between them which is known as electric force , th..."
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Blog for graduate level physics: Force on a conductor

To read an article about force on a conductor click the link given below.
Blog for graduate level physics: Force on a conductor: "We have already learned in our previous discussion that field inside a conductor is zero and the field immidiately outside is En=n(σ/ε0) ..."
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Electric field due to charged conductor

We all know that electric field inside a conductor is zero and any charge a conductor may carry lies on its surface and we have discussed it already in the earlier post (read more)but to know about the field due to this charge on the surface of the conductor you can Click the link given below
Electric field due to charged conductor
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How conductors behave in the presence of electrost...

I have written a detail article about behaviour of conductors in the presence of electric field. Please visit the link given below to read a detailed discussion about the topic.
Blog for graduate level physics: How conductors behave in the presence of electrost...: "We know that conductors like copper , silver, aluminium etc. , have very large number of free and movable charge carriers , usually one free..."
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PhysicsExpert

Electrostatic questions for board exams

Q.1 A certain region has spherical symmetry of electric field. Name the charge distribution producing such a field.

Q.2 Represent graphically the variation of electric field with distance, for a uniformly charged sphere.

Q.3 How will the radius of a flexible ring change if it is given positive charge?

Q.4 Five Charges of equal amount (q) are placed at five corners of a regular hexagon of side 20 cm. What will be the value of sixth charge placed at sixth corner of the hexagon so that the electric field at the centre of hexagon is zero ?.

Q.5 Two conducting spheres of radii r1 & r2 are at same potential. What is the ratio of charges on the spheres?

Q.7 An electric charge q is placed at one of the corner of a cube of side ‘a’. What will be the electric flux through its one of the face?

Q.8 A charge Q is divided in two parts q and Q - q separated by a distance R. If force between the two charges is maximum, find the relationship between q & Q.

Q.9 If 20 drops of same size, each having the same charge, coalesce to form a bigger drop. How will the following vary with respect to single small drop?
(i)Total charge on bigger drop
(ii) Potential on the bigger drop
(iii) The capacitance on the bigger drop
Consider the drops to have spherical symmetry.

Q.10 Work done to move a charge along a closed path inside an electric field is always zero, using this fact, prove that it is impossible to produce an electric field in which all limes of force would be parallel lines and density of their distribution would constantly increase in a direction perpendicular to the lines of force.
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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,
I_av = 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 v_d, then distance travelled by the charge carriers is Δx = v_dΔt. Putting this in equation 1 we have,
ΔQ = (nAv_dΔt)q
- Now current in the conductor is
I = ΔQ/Δt
thus,
I = nqv_dA ....................2
where v_d 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 = nqv_d
Current is a scalar quantity but current density can also be defined too include both magnitude and direction. Thus vector current density is
j = nqv_d
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 q_e is the amount of charge on a single electron and total n number of electrons passes through the wire then
I = nq_e/t
or, N = It/q_e
putting values of I , t and q_e = 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.

Capacitance IITJEE TEST Series

(A) Multiple choice questions with one or more then one correct answers

1. Which one of the following statement(s) is true
a. When the space between conductors of a capacitor is filled by dielectric it's capacitance decreases.
b. When the space between conductors of a capacitor is filled by dielectric it's capacitance increases.
c. When the space between conductors of a capacitor is filled by dielectric it's capacitance is increased by a factor known as dielectric constant of the dielectric
d. On introduction of dielectric between the conductors of capacitor , dielectric will reduce both electric field and potential difference


2 A parallel plate air capacitor is charged by cinnecting both of it's plates to the battery now a dielectric of dielectric constant k is inserted without disconnecting the battery which causes
a. Increase in potential difference across the plates and reduction in stored energy and no change in charge of plates.
b. The battery to supply the more charge to maintain the original potential difference and charge on plates increases.
c. Reduction of charge on the plates and increase of potential difference across the plates.
d None of the above.

3. For the arrangement shown below in the figure 1


total effective capacitance between the terminals is
a. 12.5 μF
b. 4.4 μF
c. 15.2 μF
d. 14 μF


4. The two plates of a capacitor are given charges ±Q and then they are immersed in a tank of oil . the electric field between the plates
a. increases
b. decreases
c. remain same
d. none of the above

ans b

5. The net capacitance of capacitors in figure 2 beetween points a and b is

a. 1 μF
b. 3μF
c. 4 μF
d 2 μF


(B) LINKED COMPREHENSION TYPE

(A) A parallel plate capacitor of area A and separation d is charged to a potential difference V and is removed from the charging source. A dielectric slab of dielectri constant k=2 , thickness d and area A/2 is inserted between the plates of the capacitor as shown below in the fig.




1. Which one of the following statement(s) is correct
a. Electric field have different values inside the dielectric as in the free space between the plates.
b. Electric field have same values inside the dielectric as in the free space between the plates.
c. There will be diferent potential difference between the portion of the plate separated by dielectric slab an between the portion separated by the vacuum.
d. none of the above.



2. If τ₁ be the surface charge density at the conductor-dielectric surface and τ₂ be the charge density at the conductor vacuum-surfacethen ratio betwen the two charge densities is
a. τ₁:τ₂=2:1
b. τ₁:τ₂=1:2
c. τ₁:τ₂=2:2
d. τ₁:τ₂=4:1



3. If C is the capacitance of the capacitor without the dielectric slab and V is the potential diffrence then capacitance and potential diffrence of capacitor on introduction of dielectric is
a. 3C/2 ; 3V/2
b. 2C/3 ; 2V/3
c. 2C/3 ; 3V/2
d. 3C/2 ; 2V/3


(B) The space between the plates of a parallel plate capacitor is filled consicutively with two dielectric layers having thickness d₁ and d₂ and permitivities ε₁ and ε₂ respectively and area of each plate is A.

1. After the introduction of dielectric slabs the capacitance of the capacitor is
a. ε₀A/[(d₁/ε₁)+(d₂/ε₁)]
b. ε₀/A[(ε₁/d₁)+(ε₂/d₂)]
c. ε₀d₁d₂/A(ε₁+ε₂)
d. ε₀A/d₁d₂(ε₁+ε₂)


2. If the voltage across the capacitor is V and electric field is directed from layer 1 to layer 2 then density of boundary charge on boundary plane is
a. (ε₂d₁-ε₁d₂)/ε₀V(ε₁-ε₂)
b. ε₀V(ε₁-ε₂)/(ε₂d₁+ε₁d₂)
c. ε₀V/(ε₂d₁-ε₁d₂)
d. ε₀V(ε₁+ε₂)/(ε₂d₁+ε₁d₂)

Capacitance II

(a) Parallel combination of capacitors-Capacitors connected in parallel combination have same potential difference across their terminals shown below in fig

where
V=potential difference across terminals of capacitors c₁, c₂, c₃
and Q₁, Q₂ and Q₃ resp. are charges on capacitors c₁, c₂, c₃.
-In parallel combination system of capacitors is equivalent to a single capacitor of capacitance
C=Q/V= C₁+C₂+C₃
where, C₁=Q₁/V
C₂=Q₂/V
C₃=Q₃/V
-Thus for capacitors connected in parallel combination their resultant capacitance C is the sum of their individual capacitances.
-Also for parallel combination of capacitors their resultant capacitance C is greater then the capacitance of greatest individual one.





(b) Series combination of capacitors
-Figure above shows three capacitors connected in series combination
-In series combination of capacitors potential difference across each capacitor is different but charge on each capacitor is same.
-The resultant capacitance of capacitors connected in series combination is the ratio of charge stored to the applied potential difference V and is given by
C=Q/V
or 1/C=V/Q
now V=V₁+V₂+V₃
=Q(1/C₁+1/C₂+1/C₃)
Thus,
1/C=1/C₁+1/C₂+1/C₃
-Resultant capacitance Cof the capacitors connected in series combination is equal to the sum of reciprocals of their individual capacitances. Here in case of series combination C is less then the capacitance of smallest individual capacitor.

(c) Energy stored in capacitor-Energy stored in capacitor is
E=QV/2
or E=CV²/2
or E=Q²/2C
factor 1/2 is due to average potential difference across the capacitor while it is charged.
-Battery supply QV amount of energy during charging a capacitor but energy stored in capacitor is QV/2 , the another half of energy is transferred into the circuit resistance in the form of heat.
thus,
heat in the wire=energy supplied by battery-energy stored in the capacitor
=QV-QV/2 = QV/2

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)

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

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=q_in/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