Question

A capacitor having $C=10\mu F$; Potential across plates, V = 10V; Distance between plates 10mm; And area of plates $10c{m}^2$, has air between the plates. What is the force on one of the plates.

• A. $4.65\times {10}^{5}N$
• B. $5.65\times {10}^{5}N$
• C. $6.65\times {10}^{5}N$
• D. None of these

Answer 1

The correct option is B $5.65\times {10}^{5}N$

Let’s first find the charge on the plate,
Q = CV
$Q=10\times {10}^{-6}\times 10$
$Q={10}^{-4}C$
Now we know that the force will simply be $\overrightarrow{F}=\overrightarrow{E}Q$
And since E is approximately constant between the plates and is given by
$E=\frac{\sigma }{2{\epsilon }_0}$
$E=\frac{Q}{2A{\epsilon }_0}$
$F=\frac{Q^2}{2A{\epsilon }_0}$
$F=\frac{({10}^{-4}{)}^2}{2\times 10\times {10}^{-4}\times 8.854\times {10}^{-12}}$
$F=5.65\times {10}^{5}N$