Question

An air filled capacitor with a dielectric slab inside it is completely charged upto energy $U$ and disconnected from battery. If the dielectric slab between the plates is removed slowly by an external agent, the work done by external agent is $3U$. The dielectric constant of the slab is:
(Assume the process to be ideal & no energy is lost)

• A. $5$
• B. $4$
• C. $3$
• D. $1.5$

Answer 1

The correct option is B $4$

Initially, the energy stored in capacitor is given by

$U=\frac{q^2}{2C}=\frac{q^2}{2K{C}_0}$

when dielectric is pulled out,

$W_{ext}=3U$

According to energy conversation;, after dielectric removed,

$U_i+W_{ext}+K{E}_i=U_f+K{E}_f$

Since slab slowly removed, means $K{E}_i=K{E}_f=0$

$\Rightarrow U+3U+0=U_f+0$

$[\because U_f=\frac{q^2}{2C_o}$, after dielectric removed]

$\Rightarrow U_f=4U$

$\Rightarrow \frac{q^2}{2C_0}=\frac{4\times q^2}{2K{C}_0}$

$\therefore K=4$

Hence, option $\left(b\right)$ is correct.