Solution:

Ans. Correct options : (c) and (d) Explanation :

(i) In case A: when the space between the plates of capacitor increases, the capacitance decreases by relation,

$$ C=\frac{K \varepsilon_{0} A}{d} $$

But battery remains same, i.e., potential difference across plate remalns ’ $V$ ’ same. So by $Q=C V$ relation, $Q$ also decreases verifies answer (c) and discardsanswer (a).

(ii) In case $B \cdot K$ is open, and capacitance decreases by moving apart plates of capacitor, so by relation $Q$ $=C V$, here $K$ is open, so charge $Q$ remains same in turn $V$ will increase on decreasing $C$. Hence answer (d) is verified.

? Very Short Answer TypeQuestion

[AT Q. 1. Does the charge given to metallic sphere depend on whether it is hollow or solid ? Give reasons for your answer.

Ans. No, Because the charge resides onty on the surface of the conductor.

Detailed Answer :

[CBSE Marking Scheme 2017]

No, the charge given to a metallic sphere does not depend on whether the metallic sphere is hollow or solid.

$1 / 2$ The amount of charge held by a capacitor is given by

$$ Q=C V $$

where, $Q=$ charge, $C=$ capacitance, $V=$ Voltage The reason that capacitance $C$ and charge $Q$ is not affected by whether or not the sphere is hollow or solid since in perfect conductor, like charges are free to take equilibrium positions in response to mutual electrostatic repulsion where charges move to outer surface of sphere and get uniformly distributed over surface of sphere. [AI Q. 1. The battery remains connected to a parallel plate capacitor and a dielectric slab is inserted between the plates. What will be effect on its (i) potential difference (ii) capacity (iii) electric field and (iv) energy stored?A [CBSE SQP I 2017]

Ans. When a battery remains connected,

(i) potential difference $V$ remains constant

(ii) capacity $C$ increases

(iii) electric field will remain same

(iv) energy stored $\frac{1}{2} C V^{2}$ increases as $C$ increases Detailed Answer :

When a battery remains connected to a parallel plate capacitor and if a dielectric slab is inserted between the plates of capacitor, then

(i) there will be no change in the potential difference as the capacitor remained connected with the battery.

(ii) capacity or capacitance will increase since with the introduction of dielectric slab, capacitance of capacitor will result $C=\frac{\kappa \varepsilon_{0} A}{d}$ where $\kappa>1$ resulting an increase in $C$. (iii) Electric field will remain same as there will be no change in potential difference and distance between the plates.

$1 / 2$

(iv) Energy stored will be increased since from the expression $U=\frac{1}{2} C V^{2}$, potential difference $V$

remains same while $C$ increases which finally increases the energy of capacitor.