Question

The energy required to charge a parallel plate condenser of plate separation $$d$$ and plate area of cross-section $$A$$ such that the uniform electric field between the plates is $$E,$$ is

A. $${ \in _0}{E^2}Ad$$  
B. $$\frac{1}{2}{ \in _0}{E^2}Ad$$
C. $$\frac{1}{2}{ \in _0}\frac{{{E^2}}}{{Ad}}$$
D. $${ \in _0}\frac{{{E^2}}}{{Ad}}$$
Answer :   $${ \in _0}{E^2}Ad$$
Solution :
Energy required to charge the capacitor is $$W = U = QV$$
$$\eqalign{ & \Rightarrow U = C{V^2} = \frac{{{\varepsilon _0}A}}{d}.{V^2} = \frac{{{\varepsilon _0}Ad}}{{{d^2}}} \cr & {V^2} = {\varepsilon _0}{E^2}Ad\,\,\left[ {\because E = \frac{V}{d}} \right]. \cr} $$

Releted MCQ Question on
Electrostatics and Magnetism >> Capacitors and Dielectrics

Releted Question 1

A parallel plate capacitor of capacitance $$C$$ is connected to a battery and is charged to a potential difference $$V.$$ Another capacitor of capacitance $$2C$$ is similarly charged to a potential difference $$2V.$$ The charging battery is now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is

A. zero
B. $$\frac{3}{2}C{V^2}$$
C. $$\frac{{25}}{6}C{V^2}$$
D. $$\frac{9}{2}C{V^2}$$
Releted Question 2

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. $$\frac{{\left( {{Q_1} - {Q_2}} \right)}}{{2C}}$$
Releted Question 3

For the circuit shown in Figure, which of the following statements is true?
Capacitors and Dielectrics mcq question image

A. With $${S_1}$$ closed $${V_1} = 15\,V,{V_2} = 20\,V$$
B. With $${S_3}$$ closed $${V_1} = {V_2} = 25\,V$$
C. With $${S_1}$$ and $${S_2}$$ closed, $${V_1} = {V_2} = 0$$
D. With $${S_1}$$ and $${S_3}$$ closed, $${V_1} = 30\,V,{V_2} = 20\,V$$
Releted Question 4

A parallel plate capacitor of area $$A,$$ plate separation $$d$$ and capacitance $$C$$ is filled with three different dielectric materials having dielectric constants $${k_1},{k_2}$$  and $${k_3}$$ as shown. 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
Capacitors and Dielectrics mcq question image

A. $$\frac{1}{K} = \frac{1}{{{K_1}}} + \frac{1}{{{K_2}}} + \frac{1}{{2{K_3}}}$$
B. $$\frac{1}{K} = \frac{1}{{{K_1} + {K_2}}} + \frac{1}{{2{K_3}}}$$
C. $$K = \frac{{{K_1}{K_2}}}{{{K_1} + {K_2}}} + 2{K_3}$$
D. $$K = {K_1} + {K_2} + 2{K_3}$$

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