*d*between the plates negligibly small compared to the length and breadth of the plates, is fully charged by connecting it across a battery of emf

*V*volt. If the capacitance of the capacitor is

*C*and the area of each plate is

*A*, the electric field at a point

*P*in between the plates, at distance

*d/*4 from the positive plate (Fig.) is

(a) *CV/*ε_{0}*A** *

(b) 4*CV/*ε_{0}*Ad*

(c) *CV/*ε_{0}*Ad*

(d) *CV/ *4ε_{0}*Ad*

(e)* *3*CV/*4ε_{0}A

The point *P* being sufficiently away from the edge of the plate, the electric field at *P* is constant and is independent of the distance from the plate. The electric field due to a plane sheet of charge is *σ*/2ε_{0} where *σ* is the surface charge density (charge per unit area). [This can be easily obtained from Gauss’s law].

In the region between the plates, the fields due to the positive plate and the negative plate act in the *same* direction ( from positive plate to negative plate) and hence they add up to produce a net field of magnitude 2×*σ*/2ε_{0} = *σ*/ε_{0}.

Since *σ* = *Q/A* and *Q = CV*, the field at *P* is *CV/*ε_{0}*A*.

[Note also that at a point outside, the electric field is zero since the positive and negative charges will produce equal and opposite fields].

(2) Two capacitors C_{1} and C_{2} are identical in all respects except for the dielectric media between their plates. C_{1} has air as dielectric where as C_{2} has a medium of dielectric constant *K* in place of air. The capacitor C_{1 }is charged to *V*_{1} volt and the charging battery is disconnected. Then the uncharged capacitor C_{2} is connected across C_{1}. If the common voltage across the capacitors is *V*_{2}, the value of the dielectric constant* K* is

(a) *V*_{1}/* V*_{2}

(b) *V*_{2}/* V*_{1}

(c) (*V*_{1 }+* V*_{2})/ (*V*_{1 }**–*** V*_{2})

(d) (*V*_{1 }**–*** V*_{2})/ *V*_{1}

(e)** **(*V*_{1 }**–*** V*_{2})/ *V*_{2}** **

If the capacitance of C_{1} is *C* the capacitance of C_{2} is *KC*.

Since the charge is conserved, the initial charge (*Q*) on C_{1} must be equal to the sum of the final charges on C_{1} and C_{2}. Therefore we have

*Q* = *CV*_{1} = *CV*_{2} + *KCV*_{2}

This gives *K =*(*V*_{1 }**–*** V*_{2})/ *V*_{2}** **

(3) Two thin metal plates of the same area are given positive charges *Q*_{1} and *Q*_{2} and kept parallel to each other to form a capacitor of capacitance *C*. If *Q*_{2} > *Q*_{1} what will be the potential difference between the plates?

(a) (*Q*_{2 }**–** *Q*_{1})_{ }/2*C*

(b) (*Q*_{2 }**+** *Q*_{1})_{ }/2*C*

(c) (*Q*_{2 }**+** *Q*_{1})_{ }/*C*

(d) (*Q*_{2 }**–** *Q*_{1})_{ }/*C*

(e)** **(*Q*_{2 }**+** *Q*_{1})_{ }/4*C* ** **

This question may mislead many of you to a wrong answer…

The potential difference (*V*) between the plates is related to the electric field (*E*) between the plates and the distance (*d*) between the plates as

*E = V/d*

The plates produce opposing fields of magnitudes *E*_{1} (due to *Q*_{1})* *and *E*_{2 }(due to *Q*_{2})* *in the region between the plates and the resultant field is directed from the plate carrying greater positive charge* Q*_{2} to that carrying smaller positive charge *Q*_{1}.

Therefore, *E* = *E*_{2 }–* E*_{1} = *V/d* so that

*V = *(*E*_{2 }–* E*_{1})*d*

But *E*_{1} = *σ*_{1}* */2ε_{0} and *E*_{2} = *σ*_{2}* */2ε_{0} where* σ*_{1} and *σ*_{2} are the surface densities of charges on the plates given by *σ*_{1} = *Q*_{1}/*A* and *σ*_{2} = *Q*_{2}/*A*

Substituting these values, *V = *[(*Q*_{2}/2ε_{0}*A*)_{ }–* *(*Q*_{1}/2ε_{0}*A*)]_{ }/*d*

Since ε_{0}*A*/*d* = *C*, we obtain *V = *(*Q*_{2 }**–** *Q*_{1})_{ }/2*C*.

(4) A parallel plate capacitor with air as dielectric has plates of area *A* and separation *d*. It is charged to a potential difference *V.* The charging battery is then disconnected and the plates are pulled apart so that the separation becomes 3*d .*

**What is the work done for pulling the plates?**

(a) ε_{0}*AV*^{2}/2*d*

(b) ε_{0}*AV*^{2}/3*d*

(c) ε_{0}*AV*^{2}/*d*

(d) 3ε_{0}*AV*^{2}/*d*

(e) 2ε_{0}*AV*^{2}/3*d*

It is enough to find the difference between the energies of the capacitor. The potential difference between the plates will increase on pulling the plates apart; but the charge will remain unchanged. The energy (*U*_{1}) before pulling is given by

*U*_{1} = *Q*^{2}/2*C* where *C* is the initial value of capacitance given by *C = *_{ }ε_{0}*A*/*d*.

When the separation between the plates is increased to 3*d*, the capacitance is reduced to *C*/3 and the energy (*U*_{2}) is given by

*U*_{2} = *Q*^{2}/(2*C/*3) = 3*Q*^{2}/2*C*

The work done (*W*) for pulling the plates apart is given by

*W = **U*_{2} – *U*_{1} = 3*Q*^{2}/2*C *– *Q*^{2}/2*C* = *Q*^{2}/*C*

* *Since *Q = CV * where *C =*_{ }ε_{0}*A*/*d*, we have

*W =** CV*^{2}* =** *ε_{0}*AV*^{2}/*d*

You will find some useful multiple choice questions with solution at physicsplus.