Question 1
A parallel plate air capacitors has plate area 0.2 m
^{2} and has separation distance 5.5 mm. Find
(a) Its capacitance when capacitor is charged to a potential difference of 500 volts
(b) Its charge
(c) Energy stored in it
(d) Force of attraction between the plates
Solution
(a) We know that for a parallel air capacitor capacitance is given by
$C = \frac {\epsilon _0 A}{d}$
Given in the question A = 0.2 m2 and d = 5.5 mm = .0055 m also we know that $\epsilon _0 = 8.854 \times 10^{-12} \; C^2/Nm^2$.
Thus,
$C = \frac {8.854 \times 10^{-12} \times 0.2 }{.0055}$
= 3.231 nF
(b) We know that charge on the capacitor is given by
Q = CV
Where V is the potential difference between plates of the capacitor which is V = 500 V.
$Q = 3.231 \times 10^{-9} \times 500$
$= 1.615 \times 10^{-6}$ C
(c) Now energy stored in the capacitor is given by
$U = \frac {CV^2}{2}$
Putting the desired values in the equation we find
$U = \frac {3.231 \times 10^{-9} \times 500 \times 500}{2}$
$= 4.04 \times 10^{-7}$ Joule
(d) Now we have to calculate the attractive force between the capacitor plates.
If F is the amount of force and d is the separation distance between the plates then work done by any person to pull the plate must be equal to the increase in energy of the system. Thus,
$W = U = Fd$
Or $F = \frac {U}{d}$
This is
$F = \frac {4.04 \times 10^{-7}} {.0055}$
$= 7.34 \times 10^{-5}$ N.
Question 2
Consider a system of capacitors where two parallel plate air capacitors each of capacitance C are connected in series to a battery of EMF ξ. Now one of the capacitor is filled uniformly with a dielectric of dielectric constant K. What would happen to electric field strength of that capacitor and what would be the change in electric field strength? Calculate the amount of charge that flows through the battery?
Solution
First we would have to calculate the charge and voltage on each capacitor. Given that capacitance of both the capacitors is same let it be C. Since both the capacitors are connected in series combination so charge on both the capacitors would be same which lead to same potential difference V across each capacitor which is
ξ = V + V or V = ξ/2
Now charge on each capacitor is
Q = CV = Cξ/2 in the absence of dielectric.
Now one of the capacitor is being filled up with dielectric of dielectric constant K. So capacitance now becomes KC, equivalent capacitance of the system now becomes
$\frac {1}{C'}=\frac {1}{KC}+ \frac {1}{C}$
Or,
$C' = \frac {KC}{1+ K}$
For finding change in electric field strength we'll calculate potential difference across the capacitor filled with the dielectric.
$V'= \frac {Q'}{KC}=\frac {KC}{2(1+ K)} \frac {\xi}{KC}= \frac {1}{2} (\frac {\xi}{1+k})$
Since V is directly proportional to electric field so as V' decreases (1\2)(1+K) times the electric field strength also decreases by the same amount.
Now flow of charge would be
$\Delta Q = Q'-Q$
$\Delta Q = \frac {KC}{2(1+ K)} \xi - \frac {C}{2} \xi=\frac {1}{2} C \xi \frac {(1-K)}{(1+K)}$
This is the required answer.
Question 3
A spherical capacitor has charges + Q and - Q on its inner and outer conductors. Find the electric potential energy stored in the capacitor?
Solution
In this problem we have to find the energy stored in a capacitor, U. We know that the spherical capacitor has capacitance
$C=\frac {4 \pi \epsilon _0 ab}{b-a}$ ---- (1)
Where a and b are the radii of the inner and outer conducting spheres. For calculating energy stored in capacitor remember the relation
$U=\frac {Q^2}{2C}$ -- (2)
Now putting the value of C from equation 1 in equation 2 we can find the energy stored in this capacitor which is
$U = \frac {Q^2}{8 \pi \epsilon _0} \frac {b-a}{ab}$
Question 4
Find the capacitance of an isolated spherical conductor of radius r
_{1 }surrounded by an adjacent concentric layer of dielectric with dielectric constant K and outside radius r
_{2}.
Solution
Let us consider that conductor in the problem has charge equals +Q Coulomb shown below in the figure.
To determine the capacitance we need to find the potential difference between conductor inside the concentric dielectric layer and the region outside the dielectric layer which is supposed to have charge -Q. Thus,
Question 5
In a parallel plate air capacitor having plate separation 0.05mm, an electric field of 4 x 10
^{4} V/m is established between the plates. After the removal of the battery a metal plate of thickness t=0.02 mm is inserted between the plates of the capacitor. Find
- Potential difference across capacitor before the introduction of metal plates.
- Potential difference across capacitor after the introduction of metal plates.
- Potential difference across capacitor if dielectric slab with dielectric constant K=3 and same thickness were inserted in place of metal plate.
Solution
Consider the figure given below
(a) Before the introduction of metal plate potential difference is
$V_0=E_0 \times d = 4 \times 10^4 \times 0.05mm = 2000$V
(b) After the introduction of metal plate let E0 be the field in the gap between the plates of the capacitor and the metal plate. Since field inside the metal plate is zero so,
$V= E_0 \times (d-t)$
Substituting the given values we get
V=1200 Volts
(c) If dielectric slab of same thickness and dielectric constant K=3 is inserted between the plates of the capacitor then
$V= E_0 \times (d-t) + \frac {E_0 t}{K}$
Substituting for the given values we get
V=1466.6 Volts.
Question 6
Find the capacitance of four parallel plates each of area Am
^{2} and separated by a distance d
_{1}, d
_{2}, and d
_{3} . The space between them is filled with dielectrics of relative permittivity ε
_{1}, ε
_{2} and ε
_{3} . Permittivity of free space is ε
_{0}.
Solution
This arrangement is equal to three parallel plate capacitors connected in series. The space between these capacitors are filled with dielectric materials having permittivity ε_{1}, ε_{2} and ε_{3} . Therefore capacitances of these three capacitors are
$C_1 = \frac {\epsilon _1 \epsilon _0A}{d_1}$ , $C_2 = \frac {\epsilon _2 \epsilon _0 A}{d_2}$ , $C_3 = \frac {\epsilon _3 \epsilon _0 A}{d_3}$
If C is the equivalent capacitance of the system then,
$\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}=\frac{1}{\varepsilon_0A}\left(\frac{d_1}{\varepsilon_1}+\frac{d_2}{\varepsilon_2}+\frac{d_3}{\varepsilon_3}\right)$
On solving above equation we find
$C=\frac{\varepsilon_0\varepsilon_1\varepsilon_2\varepsilon_3A}{\varepsilon_2\varepsilon_3d_1+\varepsilon_1\varepsilon_3d_2+\varepsilon_1\varepsilon_2d_3}$
This is the required answer.
Question 7
Consider the figure given below
- Find the equivalent capacitance between A and B
- Find the potential difference between 3μF capacitor
- Find the amount of charge on 2μF capacitor
- Find the energy stored on 12μF capacitor
Solution
(1) Consider the figure given in the question, 3μF and 2μF capacitors are connected in parallel between points A’ and B’. The equivalent capacitance between these points is
C’=3μF+2μF=5μF
Now we can assume that between the points A and B capacitors having capacitance 10 μF, 5 μF and 12 μF are connected in series as shown below in the figure
If C is the equivalent capacitance between points A and B then
$\frac{1}{C}=\frac{1}{10}+\frac{1}{12}+\frac{1}{5}=\frac{23}{60}$
or
$C=2.60 \mu$ F
(2) Charge on each of the capacitor joined in series is
q = CV = 2.6μF x 1000V = 2600μC
Therefore potential difference across points A’ and B’ i.e. across equivalent capacitor of 5μF is
V=q’/C = 2600μC/6μF = 520 Volts
Since between points A’ and B’ 2μF and 3μF capacitors are joined in parallel the potential difference across each of these will be same as between A’ and B’. Thus potential difference across 3μF capacitor is 520 Volts.
(3) Since potential difference across 2μF capacitor is 520 Volt. Hence charge on it is
q=2μF x 520 V = 1040μC
(4) The potential difference between A and B is 1000 V and equivalent capacitance is 2.602μF. Hence charge on equivalent capacitor between these points is
q=2.6μF x 1000V = 2600μC
Since between points A and B capacitors 10 μF, 5 μF and 12 μF are connected in series, the charge on each one of the capacitor is 2600μC. Thus charge on 12 μF capacitor is 2600μC
Therefore energy stored in the capacitor is
$U=\frac{q^2}{2C}=\frac{(2600\times10^{-6})^2}{2\times12\times10^{-6}}=0.28$joule
Question 8
Two parallel plate condensers A and B having capacities 2 μF and 10 μF are charged separately to the same potential of 200V. Now, positive plate of A is connected to negative plate of B and the negative plate of A is connected to positive plate of B.
- Calculate the equivalent capacitance and the common voltage of combination A-B.
- Find the loss in electrical energy in each condenser.
Solution
Consider the figure given below
(1) Let C
_{A }and C
_{B} respectively are the capacitance of condensers A and B. Initial charge on both the condensers is
q
_{A}= C
_{A}V
_{A}= (2 x 10
^{-6}) x (200V) = 4 x 10
^{-4}Coulomb
q
_{B}= C
_{B}V
_{B}= (10 x 10
^{-6}) x (200V) = 2 x 10
^{-3}Coulomb
Thus Charge on plate A
_{1}= +4 x 10
^{-4}Coulomb
Charge on plate A
_{2}= -4 x 10
^{-4}Coulomb
Charge on plate B
_{1}= -2 x 10
^{-3}Coulomb
Charge on plate B
_{2}= +2 x 10
^{-3}Coulomb
After the connection of capacitors A and B as shown above in the figure Total charge on A
_{1} and B
_{1} is
Q
_{1}=(+4 x 10
^{-4}Coulomb)+( -2 x 10
^{-3}Coulomb) = -1.6 x 10
^{-3}
And Total charge on A
_{2} and B
_{2} is
Q
_{2}=(-4 x 10
^{-4}Coulomb)+( +2 x 10
^{-3}Coulomb) = 1.6 x 10
^{-3}
Thus, after the connection both A
_{1} and B
_{1} have negative charge and both A
_{2} and B
_{2} have positive charge and magnitude of charge on all the plates is equal. Thus capacitors A and B are connected in parallel combination. Capacitance of combination of capacitor A and B is
C=C
_{A}+C
_{B} = 12 x 10
^{-6}F
Common voltage is
V=Q
_{1}/C = (1.6 x 10
^{-3})/( 12 x 10
^{-6}) = 133.3 Volts
(2) Now we have to calculate the loss in electrical energy of the capacitors
The initial energy of the system is
$E_i=\frac{1}{2}C_AV^2+\frac{1}{2}C_BV^2=.24$ joule
Energy after connection is
$E=\frac{1}{2}CV^2=.106$ joule
Therefore, loss in energy is
E
_{i}-E=.133 joule
Question 9
The plates of a parallel plate capacitor are separated by a dielectric whose relativity permittivity varies according to following relation
Find the capacitance of the capacitor.
Solution
The capacitance of parallel plate capacitor is given as
$C=\frac {\epsilon A}{d}$
Where A is the plate area, d is the separation distance and $\epsilon$ is the permittivity of the medium.
Here $\epsilon= \epsilon _r \epsilon _0$
Where $\epsilon _0$ is the permittivity of free space and
$\varepsilon_r=\frac{1}{1+\frac{x^2}{d^2}}$
Hence capacitance is
$C=\varepsilon_r\varepsilon_0\frac{A}{dx}=\frac{\varepsilon_0A}{dx}\frac{1}{1+\frac{x^2}{d^2}}$
Where dx is the variation in distance of separation as permittivity of dielectric is varying continuously with distance..
For finding the capacitance of the capacitor having continuously varying dielectric, we would have to perform integration over whole variation.
Now
$\frac{1}{C}=\frac{1}{\varepsilon_0A}{\int_{0}^{d}{\left(1+\frac{x^2}{d^2}\right)dx=\frac{1}{\varepsilon_0A}\left(x+\frac{x^3}{3d^2}\right)}}_0^d=\frac{1}{\varepsilon_0A}\left(\frac{4d^3}{3d^2}\right)=\frac{4}{3}\frac{d}{\varepsilon_0A}$
or
$C=\frac{3\varepsilon_0A}{4d}$
Question 10
Find the charge on 5 μ F capacitor in the circuit given below
Solution
The Potential Difference between AB is 6 V. Considering the branch AB, the capacitors 2 μ F and 5 μ F are in parallel and their equivalent
capacitance = 2 + 5 = 7 μ F. The branch AB then has 7 μ F and 3 μ F is series. Therefore, the effective capacitance of branch AB is
$C_{AB} = \frac {7 \times 3}{7 +3 } =\frac {21}{10}$ μ F
Total charge in branch AB, $q = C_{AB} V= \frac {21}{10} \times 6 =\frac {63}{5}$ μ F
P.D. across 3 μ F capacitor = $ \frac {q}{3} = =\frac {63}{5} \times \frac {1}{3}=\frac {21}{5}$ V
Therefore
Potential Difference across parallel combination=$6 - \frac {21}{5}=\frac {9}{5}$ V
Charge on 5 μ F capacitor= $ 5 \times 10^{-6} \times \frac {9}{5}=9$ μ C
Question 11
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.
Question 12
A Parallel Plate capacitor has following dimensions
Distance between the plates=10 cm
Area of Plate=2 m
^{2}
Charge on each plate=$8.85 \times 10^{-10}$ C
Calculate following
(a)Electric Field outside the plates
(b)Electric Field Between the plates
(c)Capacitance of the capacitor
(d)Energy stored in the capacitor
$\epsilon _0=8.854 \times 10^{-12} \ C^2N^{-1}m^{-2}$
Solution
As we know that Electric field outside the plates are zero
Electric field Inside the plates is
$E=\frac {Q}{\epsilon _0 A} =50NC^{-1}$
$C=\frac {\epsilon _0 A}{d} =17.6 \times 10^{-11}$ F
Energy stored in capacitor=$\frac {Q^2}{2C}=22.125 \times 10^{-10}$ J
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