(I) How much work does the electric field do in moving a\( - 7.7{\rm{ }}\mu C\)charge from ground to a point whose potential is\( + 65 V\)higher?

The work done by the electric field is the negative of the multiplication of the charge and the potential difference. The unit of work done can be measured in the Joules.

The given data can be listed below as:

• The value of the charge is\(q = - 7.7{\rm{ }}\mu {\rm{C}}\left( {\frac{{{{10}^{ - 6}}{\rm{ C}}}}{{1{\rm{ }}\mu {\rm{C}}}}} \right) = - 7.7 \times {10^{ - 6}}{\rm{ C }}\).
• The potential difference between the ground andthe point is \({V_{{\rm{ba}}}} = 65{\rm{ V}}\).

The electric potential difference can be expressed as:

\(\begin{aligned}{V_{{\rm{ba}}}} &= - \frac{W}{q}\\W &= - {V_{{\rm{ba}}}}q\end{aligned}\)

Here, \({V_{{\rm{ba}}}}\)is the electric potential difference between the ground (a) and the point (b) and qis the charge, and \(W\)is the work done by the electric field.

Substitute the values in the above equation.

\(\begin{aligned}W &= - 65{\rm{ V}} \times \left( { - 7.7 \times {{10}^{ - 6}}{\rm{ C}}} \right)\left( {\frac{{1{\rm{ J}}}}{{1{\rm{ C}} \cdot {\rm{V}}}}} \right)\\ &= 5 \times {10^{ - 4}}{\rm{ J}}\end{aligned}\)

Thus, the work done by the electric field is \(5 \times {10^{ - 4}}{\rm{ J}}\).