a 500 pF capacitoris charged by a 100 V battery.calculate(a) the electristatic energy stpred by it .(b) when the capacitor is disconnected from the battery and connected to another uncharged 500pF capacitor, calculate the electrostatic energy stored by the system
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Here , C = 500 x 10⁻⁶ F = 5 x 10⁻⁴ F
V = 100 V
a) Electrostatic potential energy (U) = 1/2 C(V)²
U = 1/2 x 5 x 10⁻⁴ x 100 x 100
U = 2.5 joules
Q = C V = 5 x 10⁻⁴ x 100 = 5 x 10⁻² C
b) equivalent capacitance in series = 1/500 + 1/500 = 2/500 = 1/250 = 1/C(s )
i.e. C(s) = 250 F
U = 1/2 Q²/C(s) = 1/2 x 25 x 10⁻⁴ /250 = 1/2 x 1/10 x 10⁻⁴ = 0.5 x 10⁻⁵ Joules
Here , C = 500 x 10⁻⁶ F = 5 x 10⁻⁴ F
V = 100 V
a) Electrostatic potential energy (U) = 1/2 C(V)²
U = 1/2 x 5 x 10⁻⁴ x 100 x 100
U = 2.5 joules
Q = C V = 5 x 10⁻⁴ x 100 = 5 x 10⁻² C
b) equivalent capacitance in series = 1/500 + 1/500 = 2/500 = 1/250 = 1/C(s )
i.e. C(s) = 250 F
U = 1/2 Q²/C(s) = 1/2 x 25 x 10⁻⁴ /250 = 1/2 x 1/10 x 10⁻⁴ = 0.5 x 10⁻⁵ Joules
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