Question

An air capacitor is completely charged upto the energy U and removed from battery. Now distance between plates is increased slowly by an external agent if work done by external agent is 3U then ratio of final separation between the plates to the initial separation

• A. 5
• B. 4
• C. 3
• D. 1.5

Answer 1

The correct option is B 4

initial separation $=d$

final separation $=xd$

initial capacitance $C_1=\frac{A{C}_w}{d}=1$

full sharging means $\frac{1}{2}{C}_1{v}^2=V$

or $\frac{1}{2}\frac{A{C}_w}{d}{V}^2=C$

charge $Q=C_{1}V=\frac{A{C}_w}{d}.V....\left(1\right)$

or $V=\frac{Q^2}{2C_1}....\left(2\right)$

New capacitance $=C_2=\frac{A{C}_w}{xd}=\frac{C_1}{x}$

New energy $v^{\prime }=\frac{Q^2}{2C_2}$

$=\frac{Q^2}{2\frac{C_1}{x}}=xV$

Charge remains constant

Charge in internal energy $v^{\prime }-v$

or $3v=V^{\prime }-V=V(x-1)$

$3=x-1\Rightarrow x=4$

ratio$=\frac{xd}{d}=x=4$