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.