# Chapter Notes: Capacitors Physics Class 12

Notes for Capacitor chapter of class 12 physics. Dronstudy provides free comprehensive chapterwise class 12 physics notes with proper images & diagram.

A capacitor is a device that stores electrical energy. It is an arrangement of two conductors carrying charges of equal magnitudes and opposite sign and separated by an insulating medium.

Note the following points about capacitors:
1) The net charge on the capacitor as a whole is zero. When we say that a capacitor has a charge q, we mean the positively charged conductor has a charge +q and negatively charged conductor has a charge -q.
2) The positively charged conductor is at a higher potential than the negatively charged conductor.
3) The potential difference V between the conductors is proportional to the charge q. The ratio q/V is known as capacitance C of the capacitor. Thus,

$C = {q \over V}$

4) Capacitance depends on the size and shape of the plates and the material between them. It does not depend on q or V individually.
5) The SI units of capacitance are farad(F), which is equivalent to coulomb/volt. Practical values of capacitances are usually measured in microfarad ($\mu F$).

$1\mu F = {10^{ - 6}}F$

6) It is a scalar, having dimensions

${\rm{[}}C{\rm{] }} = {\rm{ }}\left[ {{Q \over V}} \right]{\rm{ }} = {\rm{ }}\left[ {{{{Q^2}} \over W}} \right]$       $\left[ {{\rm{as }} V = {W \over Q}} \right]$
or,   $[C] = \left[ {{{{A^2}{T^2}} \over {M{L^2}{T^{ - 2}}}}} \right] = [{M^{ - 1}}{L^{ - 2}}{T^4}{A^2}]$

### Isolated Conducting Sphere as a Capacitor

A conducting sphere of radius R carrying a charge q can be treated as a capacitor. The high-potential conductor is the sphere itself and the low potential conductor is a sphere of infinite radius. The potential difference between these two sphere is

$V = {q \over {4\pi {\varepsilon _0}R}} - 0 = {q \over {4\pi {\varepsilon _0}R}}$

Hence, its capacity is (the capacitance of an isolated conductor is normally called capacity)

$C = {q \over V} = 4\pi {\varepsilon _0}R$

The capacity of a spherical conductor is directly proportional to its radius. As the potential of earth is assumed to be zero, the capacity of the earth or of any conductor connected to earth (irrespective of its shape or charge on it) will be

$C = {q \over V} = {q \over 0} = \infty$

However, if we assume the earth to be a conducting sphere of radius 6400 km, its capacitance will be,

$C = 4\pi {\varepsilon _0}R = {{6400 \times {{10}^3}} \over {9 \times {{10}^9}}} = 711{\rm{ }}\mu F$

### Energy Required to Charge a Conductor

When a conductor is charged its potential changes from 0 to V. In this process, work is done against repulsion between charge stored on the conductor and charge coming from the charging body. This work is stored as electrostatic potential energy U. So, if dq charge is given to a conductor at potential V,

$dU = dqV = dq{q \over C}$
⸫   $U = {1 \over C} \int_{{\rm{ }}0}^{{\rm{ }}{q_0}} {q dq} = {{q_0^2} \over {2C}}$  (where q0 is the total charge given to the conductor)

Since $q = CV$, we can say that the energy stored in a charged (conductor) is

$U = {1 \over 2}{{q_o^2} \over C} = {1 \over 2}{q_o}V = {1 \over 2}C{V^2}$

Note that if the charging source (say, battery) supplies charge at constant potential (say V), the work done by the charging source $W = qV$, whereas the energy stored in the charged conductor is $U = {1 \over 2}qV$. Thus, in charging a body 50 % of the energy is wasted as heat.

Application 1
Small identical droplets of distilled water (radius 0.1 mm) are found to have a charge 2 pc each. If 64 of these coalesce to form a single drop, calculate (a) the charge on it, and (b) its potential.

Solution:

(a)  From conservation of charge, we have
$Q = nq = 64 \times 2 \times {10^ - }^{12}C = {\bf{128}} \times {\bf{1}}{{\bf{0}}^ - }^{{\bf{12}}}{\bf{C}}$
(b) From conservation of mass, we have
$n \times \left( {4/3 \pi {r^3}} \right) \rho = 1 \times \left( {4/3{\rm{ }}\pi {R^3}} \right) \rho$
or,  $R = {(n)^{1/3}}r = {(64)^{1/3}} \times 0.1 \times {10^{ - 3}} = 0.4 \times {10^{ - 3}}m$
⸫   $V = {Q \over {4\pi {\varepsilon _0}R}} = {{128 \times {{10}^{ - 12}} \times 9 \times {{10}^9}} \over {0.4 \times {{10}^{ - 3}}}} = 2880$  $V$

### Sharing of Charge

Let us have two isolated spherical conductors of radii R1 and R2, charged at potentials V1 and V2.
${q_1} = {C_1}{V_1}$     and   ${q_2} = {C_2}{V_2}$
where  ${C_1} = 4\pi {\varepsilon _0}{R_1}$   and  ${C_2} = 4\pi {\varepsilon _0}{R_2}$
The combined charge is q1 + q2 and combined capacitance is C1 + C2.
Now if they are connected through a wire, charge will flow from conductor at higher potential to that at lower potential till both acquire the same potential,

$V = {{({q_1} + {q_2})} \over {({C_1} + {C_2})}} = {{{C_1}{V_1} + {C_2}{V_2}} \over {{C_1} + {C_2}}} = {{{R_1}{V_1} + {R_2}{V_2}} \over {{R_1} + {R_2}}}$ And hence, if q'1 and q'2 are the charges on the two conductors after sharing,
$q{'_1} = {C_1}V$   and    $q{'_2} = {C_2}V$   with  $(q{'_1} + q{'_2}) = \left( {{q_1} + {q_2}} \right) = q$
So,  ${{{{q'}_1}} \over {{{q'}_2}}} = {{{C_1}} \over {{C_2}}} = {{{R_1}} \over {{R_2}}}$
Thus, the charge is shared in proportion to capacity.
Some energy is lost in sharing charges. This energy is lost mainly as heat when charge flows from one body to the other through the connecting wire and also as light and sound if sparking takes place. The loss in energy is
$W = {U_I} - {U_F} = \left( {{1 \over 2}{\rm{ }}{C_1}V_1^2 + {1 \over 2}{C_2}V_2^2} \right) - {1 \over 2}\left( {{C_1} + {C_2}} \right){\rm{ }}{V^2}$
$= {{{C_1}{C_2}} \over {2({C_1} + {C_2})}}{({V_1} \sim {V_2})^2}$

Application 2
Two isolated metallic solid spheres of radius R and 2R are charged such that both of these have same charge density $\sigma$. The spheres are located far away from each other and connected by a thin conducting wire, find the new charge density on the bigger sphere.

Solution:

As charge density on both spheres is same, the total charge,

$q = {q_1} + {q_2} = 4\pi {\left( R \right)^2}\sigma + 4\pi {\left( {2R} \right)^2}\sigma = 20\pi {R^2}\sigma$   ....(i)

Now in sharing, the charge is shared in proportion to capacity (i.e., radius), so the charge on the bigger sphere,

${q'_2} = {{{R_2}} \over {({R_1} + {R_2})}}q = {{2R} \over {R + 2R}}q = {2 \over 3}q$
⸫   ${\sigma '_2} = {{{{q'}_2}} \over {4\pi {{(2R)}^2}}} = {{(2/3)q} \over {16\pi {R^2}}} = {q \over {24\pi {R^2}}} = {5 \over 6}\sigma$   [using Eqn. (i)]

### Capacitance of some Capacitors (1)  Parallel Palate Capacitor
$\sigma = {q \over A}$   and      $E = {\sigma \over {{\varepsilon _0}}} = {q \over {{\varepsilon _0}A}}$
⸫   $V = Ed = {{qd} \over {{\varepsilon _0}A}}$
⸫   $C = {q \over V} = {{{\varepsilon _0}A} \over d}$
Note that the capacitance is independent of charge given, potential raised, nature of metal or thickness of plates.

(2)  Spherical Capacitors
The induced charge q' on shell B is equal and opposite to charge q on inner sphere A.

As the charges on the two conductors are equal and opposite, the system is a capacitor.
The electric field at a point P between the shells, $E = {E_A} + {E_B} = {1 \over {4\pi {\varepsilon _0}}}{q \over {{r^2}}}$      [as ${E_B} = {E_{in}} = 0$]
or    ${{ - dV} \over {dr}} = {1 \over {4\pi {\varepsilon _0}}}{q \over {{r^2}}}$         $\left[ {{\rm{as }}E = - {{dV} \over {dr}}} \right]$
⸫   $V = - \int_{{\rm{ }}V}^{{\rm{ }}0} {} dV = {q \over {4\pi {\varepsilon _0}}}\int_{{\rm{ }}a}^{{\rm{ }}b} {} {{dr} \over {{r^2}}} = {q \over {4\pi {\varepsilon _0}}}{\rm{ }}\left[ {{1 \over a} - {1 \over b}} \right]$
⸫   $C = {q \over V} = {{4\pi {\varepsilon _0}ab} \over {[b - a]}}$

Note that
(1)  As $b \to \infty$, the capacitance reduces $4\pi {\varepsilon _0}a$. This shows that a spherical conductor is a spherical capacitor with its other plate of infinite radius.
(2)  As a and b both become very large, maintaining the difference $a - b = d$ (finite), the expression for C reduces to $C = {{{\varepsilon _0}A} \over d}$. This shows that a spherical capacitor behaves as a parallel plate capacitor if its spherical surfaces have large radii and are close to each other.

(3)  Cylindrical Capacitor : The field at a point P is
$E = {1 \over {4\pi {\varepsilon _0}}}{{2\lambda } \over r}$
But   $E = - {{dV} \over {dr}}$
⸫   $- \int\limits_v^0 {dV} = {{2\lambda } \over {4\pi {\varepsilon _0}}}\int_{{\rm{ }}a}^{{\rm{ }}b} {{{dr} \over r}}$    $\Rightarrow V = {{2\lambda } \over {4\pi {\varepsilon _0}}}{\rm{ ln}} \left( {{b \over a}} \right)$
⸫   $C = {q \over V} = {{\lambda L} \over {(\lambda /4\pi {\varepsilon _0}){\rm{ }}ln{\rm{ }}(b/a)}} = {{2\pi {\varepsilon _0}L} \over {ln{\rm{ }}(b/a)}}$

### Energy Stored in a Capacitor

If dq charge is given to a capacitor at potential V, the work done is

$dW = dq\left( V \right)$          [as $q = CV$]
or,   $W = \int_{{\rm{ }}0}^{{\rm{ }}q} {(q/C)dq} = {1 \over 2}{\rm{ }}({q^2}/C) = {1 \over 2}C{V^2} = {1 \over 2}qV$          [as $q = CV$]

This work is stored as electrical potential energy,

$U = W = C{V^2} = {{\rm{1}} \over {\rm{2}}}{{{q^2}} \over C} = {{\rm{1}} \over {\rm{2}}}qV$

This energy is not localized on the charges or the plates but is distributed in the field.
In case of a parallel plate capacitor, the field is limited between the plates, in a volume A x d. We can determine the energy density uE in this volume,

${u_{\rm{E}}} = {U \over {{\rm{Volume}}}} = {{{1 \over 2}C{V^2}} \over {Ad}} = {1 \over 2} \left[ {{{{\varepsilon _0}A} \over d}} \right] {{{V^2}} \over {Ad}}{\rm{ }}\left[ {{\rm{as }}C = {{{\varepsilon _0}A} \over d}} \right]$
${u_E} = {1 \over 2}{\varepsilon _0}{\left( {V/d} \right)^2} = {1 \over 2}{\varepsilon _0}{E^2}$       [as  ${V \over d} = E$]

### Force Between the Plates

The plates carry equal and opposite charges. There is a force of attraction between them. To calculate this force, we use the fact that electric field is conservative, for which = -(dU/dx).
In case of a parallel plate capacitor

$U = {1 \over 2}{{{q^2}} \over C} = {1 \over 2}{{{q^2}x} \over {{\varepsilon _0}A}}$                 $\left[ {{\rm{as }}C = {{{\varepsilon _0}A} \over x}} \right]$
⸫    $F = {{ - d} \over {dx}}\left[ {{1 \over 2}{{{q^2}} \over {{\varepsilon _0}A}}x} \right] = - {1 \over 2}{{{q^2}} \over {{\varepsilon _0}A}}$

The negative sign implies that the force is attractive. The force per unit area is

$\left| {{F \over A}} \right| = {1 \over 2}{{{q^2}} \over {{\varepsilon _0}{A^2}}} = {{{\sigma ^2}} \over {2{\varepsilon _0}}} = {1 \over 2}{\varepsilon _0}{E^2}$                 $\left[ {{\rm{as }}{q \over A} = \sigma \,\,\,{\rm{and}}\,\,\,E = {\sigma \over {{\varepsilon _o}}}} \right]$

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