Solving the Hard Capacitor Problem: Find the Charge

[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

Find the charge on the positive plate...it is some sort of simultaneous equations, but mine are leading me in circles. Please help!

 

[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.