Question

A small particle has charge –5.00 µC and mass 2.00 * 10^-4 kg. It moves from point A, where the electric Potential is V_A = +200 V, to point B, where the electric Potential is V_B = +800 V. The electric force is the only force acting on the particle. The particle has speed 5.00 m/s at point A. What is its speed at point B? Is it moving faster or slower at B than at A? Explain.

Short answer

Its speed is $7.416\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$at point B. It is faster at B. A negative charge gains speed when it moves to a high potential.

Step-by-step solution

Potential energy due to a point charge

Electric Potential of a charge:

$$\begin{gathered} \mathbf{V}\mathbf{=}\frac{\mathbf{U}}{{\mathbf{q}}_{\mathbf{0}}} \\ \mathbf{V}\mathbf{=}\mathbf{k}\frac{\mathbf{q}}{\mathbf{r}} \end{gathered}$$

Here U is the electric potential energy stored, q is the charge, and r is the distance traveled by the charge.

Determine the speed at point B

We know the electric potential energy due to point charge as:

$$\begin{gathered} V=\frac{U}{q} \\ =k\frac{q}{r} \end{gathered}$$

Now,

$$\begin{gathered} \mathrm{U}={\mathrm{Vq}}_0 \\ {\mathrm{q}}_0=-5\times {10}^{-6}\mathrm{C} \\ \mathrm{m}=2\times {10}^{-4}\mathrm{kg} \\ {\mathrm{v}}_{\mathrm{A}}=5\mathrm{m}/\mathrm{s} \\ {\mathrm{V}}_{\mathrm{A}}=200\mathrm{V} \\ {\mathrm{V}}_{\mathrm{B}}=800\mathrm{V} \end{gathered}$$

From the conservation of energy principle, we can write,

$K_1+U_1=K_2+U_2$

Now,

$$\begin{gathered} \frac{1}{2}{\mathrm{mv}}_{\mathrm{A}}^2+{\mathrm{V}}_{\mathrm{A}}{\mathrm{q}}_0=\frac{1}{2}{\mathrm{mv}}_{\mathrm{B}}^2+{\mathrm{V}}_{\mathrm{B}}{\mathrm{q}}_0 \\ \frac{1}{2}{\mathrm{mv}}_{\mathrm{A}}^2=\frac{1}{2}{\mathrm{mv}}_{\mathrm{A}}^2+{\mathrm{V}}_{\mathrm{A}}{\mathrm{q}}_0-{\mathrm{V}}_{\mathrm{B}}{\mathrm{q}}_0 \end{gathered}$$

Therefore, we can calculate the velocity at b as:

$v_B=\sqrt{2\times \frac{0.5m{v}_A^2+V_A{q}_0-V_B{q}_0}{m}}$

Now put the values into the equation; we get,

$$\begin{gathered} {\mathrm{v}}_{\mathrm{B}}=\sqrt{2\times \frac{0.5\times 2\times {10}^{-4}\times {\left(5\right)}^2+200\left(-5\times {10}^{-6}\right)-800\left(-5\times {10}^{-6}\right)}{2\times {10}^{-4}}} \\ =\sqrt{2\times \frac{\left(\left(2.5\times {10}^{-3}\right)+\left(-1\times {10}^{-3}\right)+4\times {10}^{-3}\right)}{2\times {10}^{-4}}} \\ =\sqrt{55} \\ =7.416\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, its speed is $\mathbf{7}\mathbf{.}\mathbf{416}\raisebox{1ex}{$\mathbf{m}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{s}$}\right.$at point B. It is faster at B. A negative charge gains speed when it moves to a high potential.