A 9500 pF capacitor holds plus and minus charges of 16.5*10-8 C. What is the voltage across the capacitor?
Q=C/V
16.5 x 10-8 / 9500 e-12
A 9500 pF capacitor holds plus and minus charges of
16.5 10 C. −8
What is the voltage across
the capacitor?
To find the voltage across the capacitor, we can use the formula:
V = Q / C
where V is the voltage, Q is the charge, and C is the capacitance.
Given:
Q = 16.5 * 10^(-8) C
C = 9500 pF = 9500 * 10^(-12) F
Let's substitute these values into the formula:
V = (16.5 * 10^(-8) C) / (9500 * 10^(-12) F)
Simplifying the formula:
V = (16.5 / 9500) * (10^(-8) / 10^(-12))
Now let's simplify the powers of 10:
V = 16.5 * (10^(-8+12)) / 9500
V = 16.5 * (10^4) / 9500
V = 173.6842 V
Therefore, the voltage across the capacitor is approximately 173.6842 V.
To find the voltage across a capacitor, we can use the formula:
V = Q / C
Where:
V is the voltage across the capacitor,
Q is the charge stored in the capacitor, and
C is the capacitance of the capacitor.
In this case, we are given the capacitance, C = 9500 pF, and the charge stored in the capacitor, Q = 16.5 * 10^(-8) C.
Let's first convert the capacitance from picofarads (pF) to farads (F) since the formula requires the capacitance to be in farads.
1 pF = 1 * 10^(-12) F (by definition)
Therefore, 9500 pF = 9500 * 10^(-12) F = 9.5 * 10^(-9) F
Now we can use the formula to find the voltage across the capacitor:
V = Q / C
= (16.5 * 10^(-8) C) / (9.5 * 10^(-9) F)
= 16.5 / 9.5 V
Calculating the result:
V = 16.5 / 9.5 V
≈ 1.7368 V
Therefore, the voltage across the capacitor is approximately 1.7368 volts.