Question

A particle with charge +4.20 µC is in a uniform electric field E directed to the left. The charge is released from rest and moves to the left; after it has moved 6.00 cm, its kinetic energy is +2.20 * 10^-6 J. What are

(a) the work done by the electric force,

(b) the potential of the starting point with respect to the end point, and

(c) the magnitude of E?

Short answer

(a) The work done by the electric force is $\mathbf{2}\mathbf{.}\mathbf{20}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{}\mathbf{J}$.

(b) The potential of the starting point with respect to the endpoint is $\mathbf{523}\mathbf{.}\mathbf{80}\mathbf{}\mathbf{V}$.

(c) The magnitude of the electric field is $\mathbf{8}\mathbf{.}\mathbf{73}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\raisebox{1ex}{$\mathbf{V}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{m}$}\right.$.

Step-by-step solution

Work done

Work done of a particle is given by,

$\frac{\mathbf{W}}{\mathbf{q}}\mathbf{=}{\mathbf{V}}_{\mathbf{a}}\mathbf{-}{\mathbf{V}}_{\mathbf{b}}$

The work done is the potential difference between two points, A and B, when the charge is a point charge.

Determine the work done by electric force

(a)

The total work done by the electric field is the kinetic energy between two points.

At the initial point, the charged particle is at rest, which means kinetic energy is zero. At the End point, kinetic energy is +2.20 × 10^-6 J.

Therefore, the work done by the electric force is $\mathbf{2}\mathbf{.}\mathbf{20}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{\hspace{0.33em}}\mathbf{J}$.

Determine the potential

(b)

The charge of the particle is 4.20 µC.

The potential is given by,

$$\begin{gathered} \frac{\mathrm{W}}{\mathrm{q}}={\mathrm{V}}_{\mathrm{a}}-{\mathrm{V}}_{\mathrm{b}} \\ {\mathrm{V}}_{\mathrm{a}}-{\mathrm{V}}_{\mathrm{b}}=\frac{\mathrm{W}}{\mathrm{q}} \\ =\frac{2.20\times {10}^{-6}\mathrm{J}}{4.20\times {10}^{-9}\mathrm{C}} \\ =5.23.80\mathrm{V} \end{gathered}$$

Therefore, the potential of the starting point with respect to the endpoint is 5.23.80 V.

Determine the magnitude of the electric field

(c)

The electric field between two points at distance l is given by,

$$\begin{gathered} V_a-V_b={\int }_0^{1}E.dl \\ {V}_a-V_b=E{\int }_0^{1}dl \\ {V}_a-V_b=El \\ E=\frac{V_a-V_b}{I} \end{gathered}$$

Now the potential difference is:${\mathrm{V}}_{\mathrm{a}}-{\mathrm{V}}_{\mathrm{b}}=523.80\mathrm{V}$ and the distance l is 0.060 m.

Put the values in the equation, and we get,

$$\begin{gathered} \mathrm{E}=\frac{{\mathrm{V}}_{\mathrm{a}}-{\mathrm{V}}_{\mathrm{b}}}{\mathrm{I}} \\ =\frac{523.80\mathrm{V}}{0.060\mathrm{m}} \\ =8.73\times {10}^3\mathrm{V}/\mathrm{m} \end{gathered}$$

Therefore, the magnitude of the electric field is $8.73\times {10}^3\mathrm{V}/\mathrm{m}$.