Capacitors A and B are two identical capacitors that store different amount of energy. A stores 6.2x 10-3J and B stores 6.8 x 10-4J. The voltage across the plates of capacitor B is 15 V. Find the voltage across the plates of capacitor A.

45.31 V.

Explanation:

From the question, since capacitor A and capacitor B are Identical, Then both Capacitor A and Capacitor B have the same Capacitance.

I.e

Ca = Cb = C

Where Ca = capacitance of Capacitor A, Cb = Capacitance of capacitor B.

Using,

Eb = 1/2C(Vb)² Equation 1

Where Eb = energy stored in capacitor B, Vb = Voltage of capacitor B

make C the subject of the equation

C = 2Eb/(Vb)² Equation 2

Given: Eb = 6.8×10⁻⁴ J, Vb = 15 V

Substitute into equation 2

C = 2(6.8×10⁻⁴)/15²

C = 6.04×10⁻⁶ F.

Similarly,

Using,

Va = √(2Ea/C) Equation 3

Where Va = Voltage Across the plates of Capacitor A, Ea = Energy stored in capacitor A.

Given: Ea = 6.2×10⁻³ J, C = 6.04×10⁻⁶ F

Substitute into equation 3

Va = √(2×6.2×10⁻³/6.04×10⁻⁶)

Va = √(2052.98)

Va = 45.31 V.

e

explanation:

there is no change in the electric potential energy as it is in rest on both conditions.

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