Two identical sheets of a metallic foil are separated by d and capacitance of the system is C and charged to E keeping the charge constant. The separation is increased by ‘l’. Then the new capacitance and potential difference will be :

$\frac{\varepsilon_0 A}{(d+l)},\left(1+\frac{l}{d}\right) E$

$q = CV = C_1 V_1$

when C = $\frac{\varepsilon_0 A}{d}$ and

$q=\frac{\varepsilon_0 A}{d}=\frac{\varepsilon_0 A}{(d+\ell)}$

∴ $\frac{\varepsilon_0 A}{d} E=\frac{\varepsilon_0 A}{(d+\ell)} E_1$

∴ $E_1 =\frac{(d+\ell)}{d} . E$

$=\left(1+\frac{\ell}{d}\right) . E$