Question

A parallel plate capacitor has $$1 \mu F$$ capacitance. One of its two plates is given $$+2 \mu C$$ charge and the other plate, $$+ 4 \mu C$$ charge. The potential difference developed across the capacitor is :-

Answer 1

Correct option is D. $$1 V$$

$$\because$$ Electric fd. due to cap plate is $$\dfrac{\sigma}{2\varepsilon _0}$$

Since both the plate has given positive charges.

$$\boxed{E=\dfrac{\sigma}{2\varepsilon_0}}$$

Net Electric .field. $$\dfrac{\sigma_2}{2\varepsilon_0}-\dfrac{\sigma_1}{2\varepsilon_0}$$ $$\boxed{\sigma=\dfrac{Q}{A}}$$

$$E_{net}=\dfrac{Q_2}{2AE_0}-\dfrac{Q_1}{2AE_0}$$

$$E_{net}=\dfrac{1}{1AE_0}(Q_2-Q_1)$$ here $$Q_2=4\mu c$$ Here

$$Q_2=2 \mu c$$

$$\boxed{E_{net}=\dfrac{1}{2AE_0}(4-2)=\dfrac{1}{AE_0}}$$

Also $$V=ED$$

$$=E_{net}\times d=\dfrac{1}{AE_0}\times d=\dfrac{1}{c}$$

$$\dfrac{1}{1}$$

$$\because C=\dfrac{E_0A}{d}$$

$$\because C=1\mu F$$

$$\boxed {\therefore V=1V}$$

$$V = \dfrac{Q}{C} = \dfrac{1\mu C}{1\mu F} = 1$$ Volt