Question

(II) The work done by an external force to move a \( - {\bf{6}}{\bf{.50}}\;{\bf{\mu C}}\) charge from point A to point B is \({\bf{15}}{\bf{.0 \times 1}}{{\bf{0}}^{{\bf{ - 4}}}}\;{\bf{J}}\). If the charge was started from rest and had \({\bf{4}}{\bf{.82 \times 1}}{{\bf{0}}^{{\bf{ - 4}}}}\;{\bf{J}}\)of kinetic energy when it reached point B, what must be the potential difference between A and B?

Short answer

The potential difference between points A and B is \( - 157\;{\rm{V}}\).

Step-by-step solution

Understanding of potential difference

When a positive test charge qmoves from point a to point b in the direction of the electric field, its potential energy changes. The difference in potential energy of the charge between points b and a is equal to the external work done in moving the charge from a to b.

\(P{E_b} - P{E_a} = - {W_{ba}}\)

The difference in potential (V) or the potential difference between points a and b is given as follows:

\({V_{\rm{b}}} - {V_{\rm{a}}} = \frac{{P{E_{\rm{b}}} - P{E_{\rm{a}}}}}{q} = \frac{{ - {W_{{\rm{ba}}}}}}{q}\)

Given information

Charge, \(q = - 6.50\;\mu {\rm{C}} = - 6.{\rm{50}} \times {\rm{1}}{{\rm{0}}^{ - 6}}\;{\rm{C}}\)

Work done by the external force,\(W' = 15.0 \times {10^{ - 4}}\;{\rm{J}}\).

The initial kinetic energy of the charge, \(K{E_{\rm{i}}} = 0\;{\rm{J}}\).

The final kinetic energy of the charge, \(K{E_{\rm{f}}} = 4.82 \times {10^{ - 4}}\;{\rm{J}}\).

Determination of total work done on the charge

If W is the work done by the electric field in moving a charge from point A to B, the potential difference between points A and B is as follows:

\(\begin{aligned}{V_{\rm{B}}} - {V_{\rm{A}}} &= \frac{{ - {W_{{\rm{BA}}}}}}{q}\\{W_{{\rm{BA}}}} &= - q\left( {{V_{\rm{B}}} - {V_{\rm{A}}}} \right)\end{aligned}\)

The total work done in moving the charge from point A to point B is the sum of work done by external force and work done by the electric force.

\(\begin{aligned}W &= W' + {W_{{\rm{BA}}}}\\ &= W' - q\left( {{V_{\rm{B}}} - {V_{\rm{A}}}} \right)\end{aligned}\)

Determination of potential difference between points A and B

According to the work-energy principle, the net work done on an object is equal to a change in its kinetic energy. So,

\(\begin{aligned}W &= \Delta KE\\W' - q\left( {{V_{\rm{B}}} - {V_{\rm{A}}}} \right) &= K{E_{\rm{f}}} - K{E_{\rm{i}}}\\{V_{\rm{B}}} - {V_{\rm{A}}} &= \frac{{W' - \left( {K{E_{\rm{f}}} - K{E_{\rm{i}}}} \right)}}{q}\end{aligned}\)

Substitute the values in the above expression.

\(\begin{aligned}{V_{\rm{B}}} - {V_{\rm{A}}} &= \frac{{15.0 \times {{10}^{ - 4}}\;{\rm{J}} - \left( {4.82 \times {{10}^{ - 4}}\;{\rm{J}} - {\rm{0 J}}} \right)}}{{ - 6.{\rm{50}} \times {\rm{1}}{{\rm{0}}^{ - 6}}\;{\rm{C}}}}\\ &= - 1.57 \times {10^2}\;{\rm{V}}\\ &= - 157\;{\rm{V}}\end{aligned}\)

Thus, the potential difference between points A and B is \( - 157\;{\rm{V}}\).