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प्रश्न
Electrostatic energy of 4 x 10−4 J is stored in a charged 25 pF capacitor. Find the charge on the capacitor.
संख्यात्मक
उत्तर
: U = 4 × 10−4 J, C = 25 pF = 25 × 10−12 F
`U = 1/2 Q^2/C`
` Q = sqrt(2UC)`
= `sqrt(2 xx 4 xx 10^-4 xx 25 xx 10^-12)`
= `sqrt(2 xx 10^-14)`
= `sqrt2 xx 10^-7 C`
= `100 sqrt2 xx 10^-9`
= `100 sqrt2 nC`
This is the charge on the capacitor.
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