A certain capacitor stores 473 J of energy when it holds 7.91 x 10−2 C of charge. What is the capacitance of this capacitor?
What is the potential difference across the plates?
Energy = 0.5C*V^2 = 473 J.
C = Q/V,
Substitute Q/V for C:
0.5(Q/V)*V^2 = 473,
0.5QV = 473,
V=473 / 0.5Q=946 / Q=946 / .0791=11960
volts.
C = Q/V = 0.0791 / 11960 = 6.6*10^-6 F.
= 6.6 uF.
To find the capacitance of the capacitor, we can use the formula:
C = Q / V
Where:
C is the capacitance (in farads, F)
Q is the charge stored in the capacitor (in coulombs, C)
V is the potential difference across the plates (in volts, V)
In this case, we are given:
Q = 7.91 x 10^-2 C
We need to find V.
To find the potential difference (V), we can use the formula to calculate the energy stored in the capacitor:
E = (1/2) * C * V^2
Where:
E is the energy stored in the capacitor (in joules, J)
C is the capacitance (in farads, F)
V is the potential difference across the plates (in volts, V)
We are given:
E = 473 J
We need to find V.
Rearranging the equation for energy, we can solve for V:
V = √(2E / C)
Substituting the given values, we have:
V = √(2 * 473 J / C)
To find the potential difference, we can first rearrange this equation to solve for C:
C = 2E / V^2
Substituting the given values, we have:
C = 2 * 473 J / V^2
Now, we can substitute this equation into the equation for capacitance:
C = Q / V
Substituting the given values, we have:
C = (7.91 x 10^-2 C) / V
Now, equate the two expressions for C:
(7.91 x 10^-2 C) / V = 2 * 473 J / V^2
Now, solve for V by cross-multiplying and rearranging the equation:
(7.91 x 10^-2 C) * V^2 = 2 * 473 J
V^2 = (2 * 473 J) / (7.91 x 10^-2 C)
V^2 ≈ 11866.997 V^2
V ≈ √(11866.997 V^2)
V ≈ 108.899 V
Therefore, the potential difference across the plates is approximately 108.899 V.
To find the capacitance, we can substitute this value back into the equation:
C = (7.91 x 10^-2 C) / (108.899 V)
C ≈ 0.000726 F
Therefore, the capacitance of this capacitor is approximately 0.000726 F.