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Two identical sheets of a metallic foil are separated by d and capacitance of the system is C and charged to a potential difference E. Keeping the charge constant, the separation is increased by ‘l’. Find the the new capacitance and potential difference .

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$q=CV=C_1V_1$
When $C= \large\frac{\in_0 A}{d}$ and
$q= \large\frac{\in_0 AE}{d} =\frac{\in_0 AE_1}{(d+l)}$
$\large\frac{\in_0 A}{d} $$E= \large\frac{\in_0 A}{(d+l)}$$E_1$
$E_1= \large\frac{(d+l)}{d}$$E= \bigg(1+\large\frac{l}{d} \bigg).E$ and $C_1= \large\frac{\in_0 A}{d+l}$
answered Jun 13, 2014 by