Question

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A 12pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor?
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Answer 1

Solution

Capacitor of the capacitance, C = 12 pF = 12 × 10 - 12 F

Potential difference, V = 50 V

Electrostatic energy stored in the capacitor is given by the relation

E = 1/2 * CV^2

= 1/2 * 12 *10^-12 * 50*50

= 1.5 * 10^-8 joule

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Answer 2

Capacitor of the Capacitance, C = 12pF

=

$12 \times 10 ^{ - 12} F \:$

Potential difference, V = 50 V

Electrostatic energy stored in the capicator is given by

E =

$\frac{1}{2}CV ^{2} = \frac{1}{2} \times 12 \times 10^{ - 12} \times (50)^{2} j \: = 1.5 \times 10 ^{ - 8} j \: \\ Therefore, the \: electrostatic \: energy \: stored \: in \: the \: capacitor is 1.5× {10}^{ - 8} J.$

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