A charge of 28.0 nC is placed in a uniform electric field that is directed vertically upward and has a magnitude of\({\bf{4}}{\bf{.00 \times 1}}{{\bf{0}}^{\bf{4}}}\;{\bf{V/m}}\). What work is done by the electric force when the charge moves (a) 0.450 m to the right; (b) 0.670 m upward; (c) 2.60 m at an angle of 45.0° downward from the horizontal?

The charge of particle in uniform electric field is\(q = 28\;{\rm{nC}}\)

The magnitude of uniform electric field is\(E = 4 \times {10^4}\;{\rm{V}}/{\rm{m}}\)

The distance moved by the electric charge is \(d = 0.450\;{\rm{m}}\)

The potential difference is calculated first then work done is calculated by multiplying electric charge in that potential difference.

The potential difference for the movement of electric charge is given below:

\(\Delta V = Ed\cos \theta \)

Here,\(\theta \)is the angle between electric field and movement of charge, its value is\(90^\circ \)because electric field is acting upward and electric charge is moving right.

Substitute all the values in the above equation.

\(\begin{aligned}\Delta V = \left( {4 \times {{10}^4}\;{\rm{V}}/{\rm{m}}} \right)\left( {0.450\;{\rm{m}}} \right)\left( {\cos 90^\circ } \right)\\\Delta V = 0\end{aligned}\)

The work done by the electric charge is given as:

\(W = q \cdot \Delta V\)

Substitute all the values in the above equation.

\(\begin{aligned}W = \left( {28\;{\rm{nC}}} \right)\left( 0 \right)\\W = 0\;{\rm{J}}\end{aligned}\)

Therefore, the work done by the electric charge is \(0\;{\rm{J}}\).