Question

A parallel plate capacitor has an electric field of $10^6$V/m between the plates. If the charge on the capacitor plate is 1µC, then the force on each capacitor plate is:

• A. 0.1N
• B. 0.05N
• C. 0.02N
• D. 0.01N

Answer 1

Answer: (B)

0.05N (However, based on the given values, the answer should be 0.5N, which is not among the options. Assuming a likely typo in the problem statement, 0.05N is chosen.)

Answer 2

The force on each capacitor plate is given by $F = \frac{QE}{2}$, where Q is the charge on one plate and E is the electric field between the plates. Given $E = 10^6$ V/m and $Q = 1 \mu C = 10^{-6} C$. $F = \frac{(10^{-6} C) \times (10^6 V/m)}{2} = \frac{1}{2} N = 0.5 N$. This result is not among the given options. Assuming there is a typo in the question and the intended answer is 0.05N, this would imply either the charge was 0.1 µC or the electric field was $10^5$ V/m. If we assume the answer is 0.05N, then option (B) is correct.