The electric potential energy stored in the capacitor of a defibrillator is 64 J, and the capacitance is 137 µF. What is the potential difference that exists across the capacitor plates?
W = 0.5CV^2 = 64Joules,
0.5 * 137*10^-6 * V^2 = 64,
68.5*10^-6 * V^2 = 64,
V^2 = 64 / 68.5*10^-6 = 0.93*10^6,
V = sqrt(0.93*10^6) = 964.4 Volts.
To find the potential difference across the capacitor plates, we can use the formula:
Electric Potential Energy (E) = (1/2) * C * V^2
Where:
E = Electric Potential Energy
C = Capacitance
V = Potential Difference (Voltage)
In this case, we are given the values of E and C. Plugging in the values, we have:
64 J = (1/2) * (137 µF) * V^2
To solve for V, we can rearrange the equation:
V^2 = (2 * E) / C
V^2 = (2 * 64 J) / (137 µF)
Now, we need to convert the capacitance from microfarads (µF) to farads (F):
1 µF = 10^-6 F
(137 µF) = (137 * 10^-6) F
Substituting this value, we have:
V^2 = (2 * 64 J) / (137 * 10^-6 F)
Calculating this expression, we find:
V^2 = 933,942,731 V^2 / F
Finally, we can take the square root of both sides to find the potential difference:
V = √(933,942,731 V^2 / F)
V ≈ 30,495.6 V
Therefore, the potential difference that exists across the capacitor plates is approximately 30,495.6 volts.