Question

-9.56 × 10⁻³ j of work is required to move 1.33 μc between points a and b. the work is done against the field. the electric potential difference between these two points is v.

Explanation:

Step1: Recall the formula for electric potential difference

The formula relating work done \( W \), charge \( q \), and electric potential difference \( \Delta V \) is \( W = q\Delta V \), so we can solve for \( \Delta V \) as \( \Delta V=\frac{W}{q} \).

Step2: Convert units to SI units

• The work done \( W = - 9.56\times10^{-3}\space J \) (the negative sign indicates work is done against the field).
• The charge \( q = 1.33\space\mu C=1.33\times 10^{- 6}\space C \) (since \( 1\space\mu C = 10^{-6}\space C \)).

Step3: Substitute values into the formula

Substitute \( W=-9.56\times 10^{-3}\space J \) and \( q = 1.33\times10^{-6}\space C \) into \( \Delta V=\frac{W}{q} \):
\[
\Delta V=\frac{- 9.56\times10^{-3}}{1.33\times10^{-6}}
\]
First, calculate the magnitude: \( \frac{9.56\times10^{-3}}{1.33\times10^{-6}}=\frac{9.56}{1.33}\times10^{-3 + 6}\approx7.188\times10^{3}=7188\space V \)
Since \( W \) is negative (work done against the field), the potential difference \( \Delta V=\frac{W}{q}=\frac{-9.56\times 10^{-3}}{1.33\times10^{-6}}\approx - 7188\space V \) (the negative sign indicates the direction of potential difference, but if we consider the magnitude of potential difference between the points, we can also report the absolute value depending on context. However, following the formula strictly, we have this result).

Answer:

\( \approx - 7188\space V \) (or if considering the magnitude of potential difference between the points, \( 7188\space V \))