Question

Figure shows two capacitors connected with a battery in a circuit.


On moving from left to right on the branch containing the capacitors, the correct graph representing potential $\left(V\right)$ vs distance $\left(x\right)$ will be:
$(C_1=3\mu \text{F};C_2=1\mu \text{F})$

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

The potential at the starting point in the branch is zero due to earthing.


As we move from left to right, the potential will increase when we cross from negative to positive plate of the capacitors $C_1$ and $C_2$

Since, both capacitors $C_1$ and $C_2$ are in series, therefore charge $q$ on them is same. So potential difference across both capacitor will be

$V_1=\frac{q}{C_1}$ and $V_2=\frac{q}{C_2}$

$\because C_1>C_2$

$\therefore V_1<V_2$

and Electric field is uniform inside capacitors so potential will increase uniformly.

$V_1=E_1{d}_1,V_2=E_2{d}_2$

$\therefore V\propto d$

Here, $E=\frac{dV}{dx}\text{(rate of change of potential})$

So as we travel across the capacitors, the rate of increase of potential will be lower for capacitor $C_1$

$V_1=\frac{q}{c_1}$

$\because C_1>C_2\Rightarrow V_1<V_{2}or{E}_1<E_2$

Thus, ${\left(\frac{dV}{dx}\right)}_1<{\left(\frac{dV}{dx}\right)}_2$

$\therefore \text{option (c) is correct graph}$