Question

The electric potential energy stored in the capacitor of a defibrillator is 73 J, and the capacitance is 120 µF. What is the potential difference that exists across the capacitor plates?​

Answer 1

Answer:

The electric potential energy stored in the capacitor of a defibrillator is 73 J, and the capacitance is 120 µF.  The potential difference that exists across the capacitor plates will be ​$1103.0261\ v$

Explanation:

• Electric potential energy (E) stored in the capacitor  can be expressed as,

            $E=\frac{1}{2} C V^{2}$    ...(1)

• Where C is the capacitance and V is the potential difference between the capacitor plates.

Rearrange equation(1) for obtaining an expression for V.

$V^{2} = \frac{2E}{C}$

Or,

$V = \sqrt{\frac{2E}{C} }$        ...(2)*

In the question, it is given that,

$E= 73\ J$

$C = 120\ \mu F = 120\times10^{-6}$F

Substitute these  values into equation (2)

$V = \sqrt{\frac{2\times73 }{120\times10^-6} } = 1103.0261\ v$