The two plates of a capacitor hold +3200 µC and -3200 µC of charge, respectively, when the potential difference is 700 V. What is the capacitance?
Cap=charge/potential 3200/700=4.57
To find the capacitance of a capacitor, we can use the formula:
C = Q / V
Where:
C is the capacitance,
Q is the charge stored in the capacitor, and
V is the potential difference across the capacitor.
In this case, the charge stored in the capacitor is given as +3200 µC and -3200 µC for the two plates, respectively. The potential difference is given as 700 V.
First, we need to convert the charge to coulombs. Since 1 µC (microcoulomb) is equal to 10^-6 C (coulombs), the charges become:
Q = +3200 µC = +3200 × 10^-6 C = +3.2 × 10^-3 C
Q = -3200 µC = -3200 × 10^-6 C = -3.2 × 10^-3 C
Now we can substitute these values into the formula:
C = Q / V
For the positive plate:
C = (+3.2 × 10^-3 C) / (700 V)
For the negative plate:
C = (-3.2 × 10^-3 C) / (700 V)
Both capacitors have the same capacitance value, regardless of the sign of the charge. Therefore, the capacitance of the capacitor is:
C = (+3.2 × 10^-3 C) / (700 V) = (-3.2 × 10^-3 C) / (700 V) = 4.57 × 10^-6 F
So, the capacitance of the capacitor is 4.57 × 10^-6 Farads.