Question

A charge of 20 µC is placed on the positive plate of an isolated parallel-plate capacitor of capacitance 10 µF. Calculate the potential difference developed between the plates.

Answer 1

The potential difference developed between the plates is 1 Volt.

Explanation:

Step 1:

Given data

Capacitance of the isolated capacitor = $-1 \cdot 0 \times 10-8$

We can arrange the above capacitance value as = $10 \mu F$

Charge across the positive plate = $20 \mu C$

Step 2:

Hence, effective charge on the capacitor = $20-0 / 2=10 \mathrm{uC}$

Step 3:

Now, the potential difference (V) between the plates = Q/C

Step 4:

By substituting the values = $10 \mathrm{uC} / 10 \times 10^{-6}$

After solving the above equation = 1V

Therefore, we can say that the potential difference developed between the plates is equal to 1 Volt.