Question

A charge of 1 µC is given to one plate of a parallel-plate capacitor of capacitance 0⋅1 µF and a charge of 2 µC is given to the other plate. Find the potential difference developed between the plates.

Answer 1

The potential difference developed between the plates is 5V

Explanation:

From the given,

Charge on the plates $\mathrm{q} 1=1 \mathrm{uC}=1 \times 10-6 \mathrm{C}$

Charge on the pates $\mathrm{q} 2=2 \mathrm{uC}=2 \times 10-6 \mathrm{C}$

To get the potential difference, we need to find the effective charge on the capacitor

$\text { Qnet }=\mathrm{Q} 2-\mathrm{Q} 1 / 2=(2-1) \times 10-6 / 2$

$\text { Qnet }=0.5 \times 10-6$

The potential difference s given by V = Qnet/C

$0.5 \times 10-6 / 0.1 \times 10-6$

After Calculations $\mathrm{V}=5 \mathrm{V}$