Physics · Alternating Current
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The energy stored in a parallel plate capacitor is 24 J. What is the potential difference be if the capacitance of the capacitor is 3 µF?
- A
4 KV
- B
16 kV
- C
54 kV
- D
8 kV
The expression relating the energy stored in a capacitor with respect to potential difference and capacitance is :
E = ½ CV²
24 = ½ (3×10^-6) V²
V² = 16 × 10⁶
V = 4 × 10³
V = 4 kV
The expression relating the energy stored in a capacitor with respect to potential difference and capacitance is :
E = ½ CV²
24 = ½ (3×10^-6) V²
V² = 16 × 10⁶
V = 4 × 10³
V = 4 kV
b) 16 kV (Kilovolts): This option suggests that the potential difference is 16 kilovolts. However, we have already calculated that the potential difference is approximately 4 kV in option (a).
c) 54 kV (Kilovolts): This option suggests that the potential difference is 54 kilovolts. Again, this does not match the calculated value of 4 kV.
d) 8 kV (Kilovolts): This option suggests that the potential difference is 8 kilovolts. However, the calculated value is approximately 4 kV as shown in option (a).
Tagged under Physics · Alternating Current · 2021