Question

−9.27×10⁻³ j of work is required to move 1.38 μc between points a and b. the work is done against the field. the electric potential difference between these two points is v. save for later submit answer

Explanation:

Step1: Recall the formula for work - potential difference relation

The formula for the work done $W$ in moving a charge $q$ through a potential difference $\Delta V$ is $W = q\Delta V$.

Step2: Rearrange the formula to solve for potential difference

We can rewrite the formula as $\Delta V=\frac{W}{q}$.

Step3: Convert the charge to SI units

The charge $q = 1.38\ \mu C=1.38\times 10^{- 6}\ C$ and the work $W=-9.27\times 10^{-3}\ J$.

Step4: Substitute the values into the formula

$\Delta V=\frac{-9.27\times 10^{-3}\ J}{1.38\times 10^{-6}\ C}$.

Step5: Calculate the potential difference

$\Delta V=\frac{-9.27}{1.38}\times10^{-3 + 6}\ V=-6.72\times10^{3}\ V$.

Answer:

$-6.72\times 10^{3}$