Capacity of a capacitor is 50 𝜇𝐹 . When it is charged from 10 mC to 50 mC, find the change in theenergy stored
Question
Capacity of a capacitor is 50 𝜇𝐹 . When it is charged from 10 mC to 50 mC, find the change in theenergy stored
Solution
The energy stored in a capacitor is given by the formula:
E = 1/2 * C * V^2
Where: E is the energy, C is the capacitance, and V is the voltage.
However, we are given the charge (Q) instead of the voltage. We can use the relationship between charge, capacitance, and voltage, which is:
Q = C * V
We can rearrange this to solve for V:
V = Q / C
We are asked to find the change in energy when the charge changes from 10 mC to 50 mC. So, we need to find the initial and final energies and subtract them.
Initial energy (E1) when Q = 10 mC:
V1 = Q1 / C = 10 * 10^-3 / 50 * 10^-6 = 200 V
E1 = 1/2 * C * V1^2 = 1/2 * 50 * 10^-6 * (200)^2 = 1 Joule
Final energy (E2) when Q = 50 mC:
V2 = Q2 / C = 50 * 10^-3 / 50 * 10^-6 = 1000 V
E2 = 1/2 * C * V2^2 = 1/2 * 50 * 10^-6 * (1000)^2 = 25 Joule
The change in energy is E2 - E1 = 25 - 1 = 24 Joule.
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