1.
Two identical metal plates are given positive charges $${Q_1}$$ and $${Q_2}\left( { < {Q_1}} \right)$$ respectively. If they are now brought close together to form a parallel plate capacitor with capacitance $$C,$$ the potential difference between them is
A.
$$\frac{{\left( {{Q_1} + {Q_2}} \right)}}{{2C}}$$
B.
$$\frac{{\left( {{Q_1} + {Q_2}} \right)}}{C}$$
C.
$$\frac{{\left( {{Q_1} - {Q_2}} \right)}}{C}$$
D.
$$\frac{{\left( {{Q_1} - {Q_2}} \right)}}{{2C}}$$
Capacitance are in series
$$\frac{1}{c} = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} \Rightarrow c = \frac{2}{3}F$$
3.
An uncharged paralle plate capacitor having a dielectric of dielectric constant $$K$$ is connected to a similar air cored parallel plate capacitor charged to a potential $${V_0}.$$ The two share the charge, and the common potential becomes $$V.$$ The dielectric constant $$K$$ is
Given, $${C_1} = {C_2} = {C_3} = 4\mu F$$
(A) The network of three capacitors can be shown as
Here, $${C_1}$$ and $${C_2}$$ are in series and the combination of two is in parallel with $${C_3}.$$
$$\eqalign{
& {C_{{\text{net}}}} = \frac{{{C_1}{C_2}}}{{{C_1} + {C_2}}} + {C_3} = \left( {\frac{{4 \times 4}}{{4 + 4}}} \right) + 4 \cr
& = 2 + 4 = 6\mu F \cr} $$
(B) The corresponding network is shown in figure below
Here, $${C_1}$$ and $${C_2}$$ are in parallel and this combination is in series with $${C_3}.$$
So, $${C_{{\text{net}}}} = \frac{{\left( {{C_1} + {C_2}} \right) \times {C_3}}}{{\left( {{C_1} + {C_2}} \right) + {C_3}}} = \frac{{\left( {4 + 4} \right) \times 4}}{{\left( {4 + 4} \right) + 4}}$$
$$ = \frac{{32}}{{12}} = \frac{8}{3}\mu F$$
(C) The corresponding network is shown below. All of three capacitors are in series.
So, $$\frac{1}{{{C_{{\text{net}}}}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} + \frac{1}{{{C_3}}} = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{3}{4}$$
$$\therefore C = \frac{4}{3}\mu F$$
(D) The corresponding network is shown below.
All of them are in parallel.
So, $${C_{{\text{net}}}} = {C_1} + {C_2} + {C_3}$$
$$ = 4 + 4 + 4 = 12\,\mu F$$
Thus, options (A) is correct.
5.
The work done in placing a charge of $$8 \times {10^{ - 18}}\,{\text{coulomb}}$$ on a condenser of capacity 100 micro-farad is
6.
A parallel-plate capacitor of area $$A,$$ plate separation $$d$$ and capacitance $$C$$ is filled with four dielectric materials having dielectric constants $${k_1},{k_2},{k_3}$$ and $${k_4}$$ as shown in the figure below. If a single dielectric material is to be used to have the same capacitance $$C$$ in this capacitor, then its dielectric constant $$k$$ is given by
7.
The capacitor, whose capacitance is $$6, 6$$ and $$3\,\mu F$$ respectively are connected in series with $$20\,volt$$ line. Find the charge on $$3\,\mu F.$$
In series $$\frac{1}{C} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} + \frac{1}{{{C_3}}}$$ and charge on each capacitor is same.
8.
A parallel plate capacitor of area $$'A'$$ plate separation $$'d'$$ is filled with two dielectrics as shown. What is the capacitance of the arrangement ?
A.
$$\frac{{3K{\varepsilon _0}A}}{{4d}}$$
B.
$$\frac{{4K{\varepsilon _0}A}}{{3d}}$$
C.
$$\frac{{\left( {K + 1} \right){\varepsilon _0}A}}{{2d}}$$
$$\therefore {c_{{\text{eq}}{\text{.}}}} = \frac{{{c_1} \times {c_2}}}{{{c_1} + {c_2}}} + {c_3} = \frac{{\left( {3 + K} \right)KA{\varepsilon _0}}}{{2d\left( {K + 1} \right)}}$$
($$\because {C_1}$$ and $${C_2}$$ are in series and resultant of these two in parallel with $${C_3}$$ )
9.
In a parallel plate capacitor, the distance between the plates is $$d$$ and potential difference across plates is $$V.$$ Energy stored per unit volume between the plates of capacitor is
A.
$$\frac{{{Q^2}}}{{2{V^2}}}$$
B.
$$\frac{1}{2}\frac{{{\varepsilon _0}{V^2}}}{{{d^2}}}$$
C.
$$\frac{1}{2}\frac{{{V^2}}}{{{\varepsilon _0}{d^2}}}$$
D.
$$\frac{1}{2}{\varepsilon _0}\frac{{{V^2}}}{d}$$
Energy stored, in parallel plate capacitor is given by $$U = \frac{1}{2}\frac{{{q^2}}}{C}$$
$$\eqalign{
& {\text{but}}\,\sigma = \frac{q}{A}\,\,\,{\text{and}}\,\,C = \frac{{{\varepsilon _0}A}}{d} \cr
& \therefore U = \frac{1}{2}\frac{{{{\left( {\sigma A} \right)}^2}}}{{\left( {\frac{{{\varepsilon _0}A}}{d}} \right)}} \cr
& {\text{or}}\,\,U = \frac{{A{\sigma ^2}d}}{{2{\varepsilon _0}}} \cr
& {\text{or}}\,\,U = \frac{1}{2}{\left( {\frac{\sigma }{{{\varepsilon _0}}}} \right)^2} \times {\varepsilon _0}Ad \cr
& {\text{or}}\,\,U = \frac{1}{2}{E^2}{\varepsilon _0}Ad \cr} $$
Energy stored per unit volume i.e. energy density is thus given by
$$\eqalign{
& u = \frac{U}{V} = \frac{U}{{Ad}} = \frac{1}{2}{\varepsilon _0}{E^2} \cr
& \Rightarrow u = \frac{1}{2}{\varepsilon _0}{\left( {\frac{V}{d}} \right)^2} = \frac{1}{2}{\varepsilon _0}\frac{{{V^2}}}{{{d^2}}} \cr} $$
10.
A disc of radius $$\frac{a}{4}$$ having a uniformly distributed charge $$6C$$ is placed in the $$x - y$$ plane with its centre at $$\left( { - \frac{a}{2},0,0} \right).$$ A rod of length $$a$$ carrying a uniformly distributed charge $$8C$$ is placed on the $$x$$-axis from $$x = \frac{a}{4}$$ to $$x = \frac{{5a}}{4}.$$ Two point charges $$ - 7 C$$ and $$3 C$$ are placed at $$\left( {\frac{a}{4}, - \frac{a}{4},0} \right)$$ and $$\left( { - \frac{{3a}}{4},\frac{{3a}}{4},0} \right)$$ respectively. Consider a cubical surface formed by six surfaces $$x = \pm \frac{a}{2},y = \pm \frac{a}{2},z = \pm \frac{a}{2}.$$ The electric flux through this cubical surface is
From the figure it is clear that the charge enclosed in the cubical surface is $$3C + 2C - 7C = - 2C.$$ Therefore the electric flux through the cube is