Question

A condenser $\text{A}$ has a capacity of $15\mu \text{F}$ when it is filled with a medium of dielectric constant $15$. Another condenser $\text{B}$ has a capacity of $1\mu \text{F}$ with air between the plates. Both are charged separately by a battery of $100\text{V}$. After charging, both are connected in parallel without the battery and the dielectric medium is removed. The common potential now is

• A. $400\text{V}$
• B. $800\text{V}$
• C. $1200\text{V}$
• D. $1600\text{V}$

Answer 1

The correct option is B $800\text{V}$

Given that,
Capacitance of the condenser $\text{A}$, $C_1=15\mu \text{F}$

Dielectric constant of the medium in condenser $\text{A}$, $K=15$

Capacitance of the condenser $\text{B}$, $C_1=15\mu \text{F}$

Initial potential difference across the condensers, $V=100\text{V}$,

As battery is disconnected , the charges on two capacitors remain constant . The charges on the capacitors are

$Q_1=15\times 100=1500\mu \text{C}$ and
$Q_2=1\times 100=100\mu \text{C}$

As dielectric is removed so the capacitance $15\mu \text{F}$ becomes ,

$C_1^{\prime }=\frac{C_1}{K}=\frac{15}{15}=1\mu \text{F}$

Capacitance $1\mu \text{F}$ will be same $C_2^{\prime }=C_2=1\mu \text{F}$

So, the new common potential,

$V^{\prime }=\frac{Q_1+Q_2}{C_1^{\prime }+C_2^{\prime }}=\frac{1500+100}{1+1}$

$\therefore V^{\prime }=800\text{V}$

Hence, option (d) is the correct answer.