Question

Two condensers of capacity C and $2$C are connected in parallel and these are charged upto V volt. If the battery is removed and dielectric medium of constant K is put between the plates of first condenser, then the potential at each condenser is

• A. $\frac{V}{k+2}$
• B. $2+\frac{k}{3V}$
• C. $\frac{2V}{k+2}$
• D. $\frac{3V}{k+2}$

Answer 1

Answer: (D)

$\frac{3V}{k+2}$

total charge $Q=3CV.$

Now when battery is disconnected charge remains constant on the plates.

And when dielectric is added capacitance become kc,

Now let charge be x on the capacitor. (total charge is Q)

We know $V=\frac{Q}{C}$

Now voltage difference is same across the capacitor

$\therefore \frac{x}{2c}=\frac{Q-x}{kc}\Rightarrow kx=2Q-2x$

$x=\frac{2Q}{2+k}$

$\therefore$ charge across $2$c capacitor $=\frac{2Q}{2+k}$

$\therefore VoltageV=\frac{2Q}{(2+k)\left(2c\right)}=\frac{Q}{c(2+k)}=\frac{3cV}{c(2+k)}=\frac{3V}{2+k}$