The charge on one plate of a capacitor is + 60 nC and the charge on the other plate is - 60 nC. If the potential difference between the plates is 150 Volts, what is the capacitance of the capacitor?
To find the capacitance of a capacitor, we can use the formula:
C = Q / V
Where C is the capacitance, Q is the charge on one of the plates, and V is the potential difference between the plates.
In this case, the charge on one plate is +60 nC and the charge on the other plate is -60 nC. Since the charge on one plate is positive and the charge on the other plate is negative, the total charge (Q) is the sum of the absolute values of these charges:
Q = |60 nC| + |-60 nC| = 60 nC + 60 nC = 120 nC
The potential difference (V) between the plates is given as 150 Volts.
Now we can substitute these values into the formula to find the capacitance:
C = Q / V = 120 nC / 150 V
To simplify the units, we need to convert nC to Coulombs (C), since the SI unit for charge is Coulombs. We know that 1 nC = 10^-9 C, so we can convert the charge:
Q = 120 nC = 120 * 10^-9 C = 1.2 * 10^-7 C
Substituting this value and the potential difference into the formula:
C = (1.2 * 10^-7 C) / 150 V
Now we can calculate the capacitance:
C ≈ 8 * 10^-10 F (approximately)
Therefore, the capacitance of the capacitor is approximately 8 * 10^-10 Farads.