What voltage is required to store 7.0 10-5 C of charge on the plates of a 5.0 µF capacitor?
_________ V
Q = C V
V = Q/C
C is the capacitance in Farads
Q is the charge in Coulombs
71428.57 coulomb
First class
12
V=Q/C
=12
Why did the capacitor go to the gym?
To get charged up, of course!
Now, let's calculate the voltage. We can use the formula Q = CV, where Q is the charge, C is the capacitance, and V is the voltage.
Plugging in the values, we have:
7.0 * 10^-5 C = (5.0 * 10^-6 F) * V
Now, let's solve for V:
V = (7.0 * 10^-5 C) / (5.0 * 10^-6 F)
V ≈ 14 V
So, the voltage required to store 7.0 * 10^-5 C of charge on the plates of a 5.0 µF capacitor is approximately 14 V.
To find the voltage required to store a certain amount of charge on a capacitor, you can use the formula:
V = Q / C
where V is the voltage, Q is the charge, and C is the capacitance.
In this case, the charge is given as 7.0 * 10^(-5) C and the capacitance is given as 5.0 µF (which is equivalent to 5.0 * 10^(-6) F).
Substituting these values into the formula, we have:
V = (7.0 * 10^(-5) C) / (5.0 * 10^(-6) F)
To divide two values in scientific notation, you subtract the exponents:
V = 7.0 * 10^(-5 - (-6)) / F
Simplifying the exponents:
V = 7.0 * 10^(-5 + 6) / F
V = 7.0 * 10^(1) F
Multiplying 7.0 by 10^1:
V = 7.0 * 10 / F
Dividing 70 by 5.0:
V = 14 / F
The voltage required to store 7.0 * 10^(-5) C of charge on a 5.0 µF capacitor is 14 V.