Question: Q. 6. (i) How many electrons must be added to one plate and removed from other so as to store $25.0 \mathrm{~J}$ of energy in a $5.0 n \mathrm{~F}$ parallel plate capacitor ?
(ii) How would you modify this capacitor so that it can store $50.0 \mathrm{~J}$ of energy without changing the charge on its plates?
U] [SQP 2017-18]
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Solution:
Ans. (i)
$$ \begin{aligned} & C=5 \times 10^{-9} F, U=25 \mathrm{~J} \ & U=\frac{Q^{2}}{2 C} \ & Q^{2}=2 U C \ & =2 \times 25 \times 5 \times 10^{-9} \ & Q=5.0 \times 10^{-4} C \ & Q=n e \ & n=\frac{Q}{e}=\frac{5 \cdot 0 \times 10^{-4} C}{1.60 \times 10^{-19}} \ & =3 \cdot 125 \times 10^{15} \text { electrons } 1 / 2 \end{aligned} $$
(ii) Without changing charge on the plates, we can make $C$ half.
$$ C=\frac{\varepsilon_{0} A}{d} $$
Double the plate separation or inserting dielectric of dielectric constant of a value such that $C$ will become half.
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CBSE Marking Scheme 2017]
