91Ó°ÊÓ

By the work energy theorem,

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

$\begin{array}{r}{\mathrm{W}}_{\text{tot}}=\mathrm{Δ°­}\\ ={\mathrm{K}}_b−K_a\\ =4.35×{10}^{−5}\mathrm{J}\end{array}$

Refer to the image below,

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

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

Thus, the work done by the electric force alone is $-2.15×{10}^{-5} J$.

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,

$\begin{array}{r}{W}_{F_{\mathrm{Ó\copyright }}}=q\left(V_a−V_b\right)\\ \left(V_a−V_b\right)=\frac{W_{F_{\mathrm{Ó\copyright }}}}{q}\\ =\frac{−2.15×{10}^{−5}\mathrm{J}}{7â‹\dots 60×{10}^{−9}\mathrm{C}}\\ =−2.83×{10}^3\mathrm{V}\end{array}$

So, the standing point is a and the potential at this point is $2.83×{10}^3 V$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,

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

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