# Hard Capacitor problem

1. Feb 19, 2004

### Matt Jacques

Although I figure you will say easy :P

What we know:

• Area of plates: 5 meters squared
• When the plates are separated .004 m, the voltage increases 100V

What we don't know:

• The original plate separation
• The original voltage
• The capacitance

2. Feb 20, 2004

### NateTG

The voltage in a capacitor is:
$$V=\frac{Q}{C}$$
and
$$C=\frac{\epsilon_0 A}{d}$$

so you've got
$$V=\frac{Qd}{\epsilon_0A}$$
so
$$V_0-V_1=\frac{Qd_0}{\epsilon_0A}-\frac{Qd_1}{\epsilon_0A}$$
The change in voltage is 100, the area is 5, and $$\epsilon_0$$ is a constant. It doesn't look so bad to me.