91影视

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 \(0\;{\rm{J}}\).