24.
Study the following paragraph and answer question numbers (A) to (D) based on it.
The capacity of a capacitor increases both, when a conducting slab or an insulating slab is introduced between the plates of the capacitor. In the former case, electric field $E=0$ inside the conductor and in the latter case, $E < E_{0}$, inside the insulator. Thus, potential difference $V=E\times d$ decreases and hence capacity $C=Q/V$ increases. It should be clearly understood that when a dielectric slab is introduced in between the plates of a charged capacitor with battery connected across the plates, (i) Capacity (C) and charge (Q) increases. (ii) Potential V remains constant. (iii) Energy stored $U=\frac{1}{2}CV^{2}$ increases. (iv) Electric field decreases. The electric energy density in a region with electric field is $\frac{1}{2}\epsilon_{0}E^{2}$.
(A) When a number of capacitors are connected in parallel between two points, the equivalent capacitance?
(a) increases
(b) decreases
(c) remain the same
(d) None of the above.
(B) A parallel plate capacitor is charged. If the plates are pulled apart?
(a) the capacitance increases
(b) the total charge increases
(c) the potential increases
(d) the charge and the potential difference remains the same.
(C) If there are n capacitors in parallel connected to V volt source, then the energy stored is equal to:
(a) CV
(b) $\frac{1}{2}n~CV^{2}$
(c) $CV^{2}$
(d) $\frac{1}{2n}CV^{2}$
(D) A 15 pF capacitor is connected to a 100V battery. The electrostatic energy stored in the capacitor is:
(a) $1.5\times10^{-8}J$
(b) $15\times10^{-8}J$
(c) $75\times10^{-8}J$
(d) $7.5\times10^{-8}J$