Question 262: Electrostats & Capacitance

Question

A capacitor with capacitance 5μF is charged to
5μC. If the plates are pulled apart to reduce the
capacitance to 2μF, how much work is done ?

Accepted Answer

Work done = $$\Delta $$U
= U
f
Ã¢ÂÂ U
i
$$ = {{{q^2}} \over {2{C_r}}} - {{{q^2}} \over {2{C_i}}}$$
$$ = {{{{\left( {5 	imes {{10}^{ - 6}}} ight)}^2}} \over 2}.\left( {{1 \over {2 	imes {{10}^{ - 6}}}} - {1 \over {5 	imes {{10}^{ - 6}}}}} ight)$$
$$ = {{15} \over 4} 	imes {10^{ - 6}} = 3.75{m{ }} 	imes {m{ }}{10^{-6}}{m{ }}J$$

Answer

2.16 × 10–6 J

Answer

2.55 × 10–6 J

Answer

3.75 × 10–6 J

Answer

6.25 × 10–6 J

Electrostats & Capacitance

A capacitor with capacitance 5μF is charged to 5μC. If the plates are pulled apart to reduce the capacitance to 2μF, how much work is done ?

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