A 5.0 µC point charge is moved within an electric field and has an electric potential energy change of 10.0 J. What is the electric potential difference before and after the charge was moved? Show work. (µC = 1.0 × 10–6 C)
Vq=energy
so delta V must be 10 volts.
To find the electric potential difference, we need to use the formula:
ΔPE = qΔV
Where:
ΔPE is the change in electric potential energy
q is the charge
ΔV is the electric potential difference
We are given:
ΔPE = 10.0 J
q = 5.0 µC = 5.0 × 10^(-6) C
Rearranging the formula, we have:
ΔV = ΔPE / q
Substituting the given values:
ΔV = 10.0 J / (5.0 × 10^(-6) C)
To simplify the calculation, we can convert the charge to Coulombs:
5.0 µC = 5.0 × 10^(-6) C
Now, let's calculate the electric potential difference:
ΔV = 10.0 J / (5.0 × 10^(-6) C)
ΔV = 2.0 × 10^6 J/C
Therefore, the electric potential difference before and after the charge was moved is 2.0 × 10^6 J/C.