A parallel plate capacitor has plates of area `200cm^2 and separation between the plates 1.00 mm. What potential difference will be developed if a charge of 1.00 nC )(i.e., 1.00X10^(-9) C) is given to the capacitor? If the plate separation is now increased to 2.00 mm, what will be the new potential difference?

Explanation

To solve the problem step by step, we will first calculate the capacitance of the parallel plate capacitor, then use it to find the potential difference for the given charge. Finally, we will recalculate the potential difference after increasing the plate separation.

### Step 1: Calculate the Capacitance of the Capacitor

The formula for the capacitance \( C \) of a parallel plate capacitor is given by:

\[
C = \frac{\varepsilon_0 A}{d}
...