A particle with charge +7.60 nC is in a uniform electric field directed to the left. Another force, in addition to the electric force, acts on the particle so that when it is released from rest, it moves to the right. After it has moved 8.00 cm, the additional force has done 6.50 * 10-5 J of work and the particle has 4.35 * 10-5 J of kinetic energy. (a) What work was done by the electric force? (b) What is the potential of the starting point with respect to the end point? (c) What is the magnitude of the electric field?

By the work energy theorem,

Total work done= change in kinetic energy in the system.

$\begin{array}{r}{\mathrm{W}}_{\text{tot}}=\mathrm{\Delta K}\\ ={\mathrm{K}}_b-K_a\\ =4.35\times {10}^{-5}\mathrm{J}\end{array}$

Refer to the image below,

The total work done is due to both the additional force F and the electric force F_E,

$\begin{array}{r}{W}_{\text{tot}}=W_{F_z}+W_F\\ W_{F_z}=W_{\text{tot}}-W_F\\ =4.35\times {10}^{-5}\mathrm{J}-6.50\times {10}^{-5}\mathrm{J}\\ =-2.15\times {10}^{-5}\mathrm{J}\end{array}$

Thus, the work done by the electric force alone is $-2.15\times {10}^{-5}\hspace{0.33em}J$.

In this diagram,

The electric force acts in the left direction since the charge is positive and the displacement is in the right direction. So, the electric force does negative work while the additional force produces positive work by the displacing the charge in the right direction.

Now, for the electric force,

$\begin{array}{r}{W}_{F_{\mathrm{\Xi }}}=q\left(V_a-V_b\right)\\ \left(V_a-V_b\right)=\frac{W_{F_{\mathrm{\Xi }}}}{q}\\ =\frac{-2.15\times {10}^{-5}\mathrm{J}}{7\cdot 60\times {10}^{-9}\mathrm{C}}\\ =-2.83\times {10}^3\mathrm{V}\end{array}$

So, the standing point is a and the potential at this point is $2.83\times {10}^3\hspace{0.33em}V$below the potential at point b.

The Work done and electric field relation is,

${\mathrm{W}}_{F_E}=-{\mathrm{F}}_E\mathrm{d}=-\mathrm{qEd}$

Solve for the electric field,

$\begin{array}{r}E=-\frac{V_a-V_b}{d}\\ =-\frac{-2.83\times {10}^3\mathrm{V}}{0.0800\mathrm{m}}\\ =3.54\times {10}^4\mathrm{V}/\mathrm{m}\end{array}$

Thus, the electric field magnitude is$3.54\times {10}^4\hspace{0.33em}\mathrm{V}/\mathrm{m}$ .