Question

The capacity of a parallel plate condenser is $10\mu F$ without the dielectric. Material with a dielectric constant of $2$ is used to fill half-thickness between the plates. The new capacitance is $\underset{–}{}\mu F$ :

• A. $10$
• B. $13.33$
• C. $15$
• D. $20$

Answer 1

Answer: (B)

Initial capacitance $=\frac{{ϵ}_{0}A}{d}=10\mu F$
Now effective capacitance is series combination of $\frac{{ϵ}_{0}A}{\frac{d}{2}}$ and $\frac{k{ϵ}_{0}A}{\frac{d}{2}}$
$\therefore C_{eff}=$ series of $20\mu F$ and $40\mu F$
$\frac{1}{C_{eff}}=\frac{1}{20}+\frac{1}{40}$
$C_{eff}=13.33\mu F$