Question

In the given figure, a capacitor of non-parallel plates is shown. The plates of capacitor are connected by a cell of emf $V_0$. If $\sigma$ denotes surface charge density and $E$ denotes electric field. Then

• A. $E_F=E_D$
• B. ${\sigma }_A>{\sigma }_B$
• C. ${\sigma }_A={\sigma }_B$
• D. $E_F>E_D$

Answer 1

The correct option is B ${\sigma }_A>{\sigma }_B$

We know that,

$E=\frac{V_0}{d}$
$(E=\text{Pelectric field})$

Where, $V_0$ and $d$ is the potential difference and perpendicular distance between two plates at the considered point.

Here, potential difference between plates at point $D$ and$F$ are same but distance $d$ between plates is more at $F$ as compared to that at $D$

$E_D>E_F$

$\sigma =$ surface charge density

$E_D=\frac{{\sigma }_A}{{\epsilon }_0}$

$E_F=\frac{{\sigma }_B}{{\epsilon }_0}$

But $E_D>E_F$ $so,{\sigma }_A>{\sigma }_8$
Hence, option (a) is correct.