The electric potential energy stored in the capacitor of a defibrillator is 77 J, and the capacitance is 123 µF. What is the potential difference that exists across the capacitor plates?
PE=CU²/2
U=sqrt{2•PE/C}
To find the potential difference across the capacitor plates, we can use the equation:
Electric potential energy (E) = 1/2 * C * V^2
Where:
- E is the electric potential energy stored in the capacitor
- C is the capacitance (in Farads)
- V is the potential difference (in volts)
In this case, we are given the electric potential energy (E = 77 J) and the capacitance (C = 123 µF).
First, we need to ensure that the capacitance is in Farads. The given value is in microfarads (µF), so we need to convert it to Farads by dividing it by 10^6:
C = 123 µF = 123 * 10^(-6) F = 0.000123 F
Now we can rearrange the equation for electric potential energy and solve for the potential difference (V):
E = 1/2 * C * V^2
V^2 = 2E/C (multiply both sides by 2/C)
V = √(2E/C) (take the square root of both sides)
Substituting the given values:
V = √(2 * 77 J / 0.000123 F)
V = √(154000 J/F)
V ≈ 393 V
Therefore, the potential difference that exists across the capacitor plates is approximately 393 volts.