Two uncharged metal spheres, spaced 20.0 cm apart, have a capacitance of 28.0 pF.
How much work would it take to move 14.0 nC of charge from one sphere to the other?
*Please report answer in Joules.
To calculate the work required to move charge between the spheres, we can use the formula:
W = 1/2 * C * V^2
Where:
W is the work done
C is the capacitance
V is the potential difference between the spheres
First, we need to calculate the potential difference between the spheres. The capacitance formula can be rearranged as follows:
C = q / V
Where:
C is the capacitance
q is the charge
V is the potential difference between the spheres
Rearranging this formula to solve for V:
V = q / C
Now we can substitute the given values to calculate the potential difference:
q = 14.0 nC = 14.0 x 10^(-9) C
C = 28.0 pF = 28.0 x 10^(-12) F
V = (14.0 x 10^(-9) C) / (28.0 x 10^(-12) F)
= 500 V
Now we can substitute the potential difference into the work formula:
W = 1/2 * (28.0 x 10^(-12) F) * (500 V)^2
= 1/2 * (28.0 x 10^(-12) F) * (250,000 V^2)
= 3.5 x 10^(-6) J
Therefore, it would take 3.5 x 10^(-6) Joules of work to move 14.0 nC of charge from one sphere to the other.
To calculate the work required to move charge between the spheres, we need to use the formula:
W = (1/2) * C * V^2,
where W is the work done, C is the capacitance, and V is the potential difference across the spheres.
First, let's find the potential difference. The potential difference between the spheres is given by:
V = q / C,
where q is the charge. In this case, q = 14.0 nC (nanocoulombs). Let's convert it to coulombs:
q = 14.0 nC = 14.0 * 10^-9 C.
Now, we can calculate the potential difference:
V = (14.0 * 10^-9 C) / (28.0 * 10^-12 F) = 0.5 V.
Now that we have the potential difference, we can calculate the work done:
W = (1/2) * (28.0 * 10^-12 F) * (0.5 V)^2.
Simplifying this equation, we get:
W = (1/2) * 28.0 * 10^-12 * (0.5)^2 = 0.175 * 10^-12 J,
which can be written as:
W = 1.75 * 10^-14 J.
Therefore, it would take approximately 1.75 * 10^-14 Joules of work to move 14.0 nC of charge from one sphere to the other.
A voltage difference of V = Q/C = 14*10^-9/28*10^-12 = 500 V will appear between the spheres after the transfer is made.
The electrical energy of the charged spheres is (1/2) C V^2
You don't need to know or use the separation distance.
It's a pretty low number: microjoules