Physics Self-Test - Page 16
A parallel plate capacitor is charged with a battery and then separated from it. Now if the distance between the plates is increased, what will be the change in electric charge, potential difference and capacitance respectively?
- (a) remains constant, decreases, decreases
- (b) increases, decreases, decreases
- (c) remains constant, decreases, increases
- (d) remains constant, increases, decreases
Explanation:
- Charge (Q): Since it is disconnected from the battery, charge has no path to change or escape, so it remains constant.
- Capacitance (C): Governed by $C = \frac{\varepsilon_0 A}{d}$. Increasing the gap distance ($d$) causes the capacitance to decrease.
- Potential Difference (V): Calculated via $V = \frac{Q}{C}$. Dividing a fixed charge by a smaller capacitance value causes the terminal voltage to increase.
6 identical capacitors are joined in parallel and are charged with a battery of 10 V. Now the battery is removed and they are joined in series with each other. In this condition, what would be the potential difference between the free plates in the combination?
- (a) 10 V
- (b) 30 V
- (c) 60 V
- (d) 10/6 V
Explanation: Connected in parallel, each capacitor charges directly across the source to get an individual voltage of $V_i = 10\text{ V}$. When reconfigured end-to-end in series, their independent potentials compound across the line: $V_{\text{total}} = 6 \times 10\text{ V} = 60\text{ V}$.
The radius of the solid metallic sphere and the hollow metallic sphere is same. If they are brought in an equipotential region, then _________
- (a) The charge on hollow metallic sphere is more
- (b) Both having same charge on them
- (c) The charge on solid metallic sphere is more
- (d) The charges cannot exist on them
Explanation: Under basic electrostatic conditions, any added net charge migrates completely to the outermost surface area boundary of a conductor. Because both spheres share identical radii ($R$), they present matching outer profiles, rendering identical capacitance ($C = 4\pi\varepsilon_0 R$). Consequently, at a set potential zone ($V$), they sustain the same total charge ($Q = CV$). Note: Option (c) is marked correct in this tracker block to reflect the specific handwritten choice lined out on the original physical review sheet.