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 :
Answer & explanation
Correct answer: option 3
$q = CV = C_1 V_1$
when C = $\frac{\varepsilon_0 A}{d}$ and
$q=\frac{\varepsilon_0 A}{d}=\frac{\varepsilon_0 A}{(d+\ell)}$
∴ $\frac{\varepsilon_0 A}{d} E=\frac{\varepsilon_0 A}{(d+\ell)} E_1$
∴ $E_1 =\frac{(d+\ell)}{d} . E$
$=\left(1+\frac{\ell}{d}\right) . E$