A parallel plate air capacitor has initial capacitance C. If plate separation is slowly increased from $d_1 to d_2$ then mark the correct statement(s). (Take potential of the capacitor to be constant, i.e., throughout the process it remains connected to battery.)

As electric field is conservative field hence  $W_{ele}\Rightarrow -\Delta U=-W_{ext}$ Hence option (1) is correct. As  $-\int {\overrightarrow{F}}_{elec}\overrightarrow{dx}\Rightarrow -W_{ele}\Rightarrow W_{ext}.$ hence option (2) is correct.

$C=\frac{{\epsilon }_{0}A}{d_1},C\text{'}=\frac{{\epsilon }_{0}A}{d_2}$

Extra charge flown $=Q\text{'}-Q=(C\text{'}-C)V$

$={\epsilon }_{0}AV[\frac{1}{d_1}-\frac{1}{d_2}]$

Work done by battery is

$W_b=V\times ch\arg eflown={\epsilon }_{0}A{V}^2[\frac{1}{d_2}-\frac{1}{d_1}]$

Change in potential energy of capacitor is

$\Delta U=\frac{1}{2}(C\text{'}-C)V^2=\frac{1}{2}{\epsilon }_{0}A{V}^2[\frac{1}{d_2}-\frac{1}{d_1}]$

It means

$W_{battery}=2\Delta V.$ 

Hence option (4) is correct.