what is the electric potential of a 4.5×10⁻⁵ c charge that has an electric potential energy of 0.027 j?

Explanation:

Step1: Recall the formula

The formula for electric - potential energy is $U = qV$, where $U$ is the electric - potential energy, $q$ is the charge, and $V$ is the electric potential. We need to solve for $V$.

Step2: Rearrange the formula

Rearranging $U = qV$ for $V$ gives $V=\frac{U}{q}$.

Step3: Substitute the values

Given $U = 0.027\ J$ and $q = 4.5\times10^{-5}\ C$. Substitute these values into the formula: $V=\frac{0.027}{4.5\times10^{-5}}$.

Step4: Calculate the result

$\frac{0.027}{4.5\times10^{-5}}=\frac{2.7\times10^{-2}}{4.5\times10^{-5}}=\frac{2.7}{4.5}\times10^{3}=0.6\times10^{3} = 600\ V$.

Answer:

600