91Ó°ÊÓ

The potential energy of a charge is given by:

U=qV

Here q is the charge, and V is the potential of the charge q.

(a)

The potential energy of a charge is given by:

U = qV

And the kinetic energy is

${\mathrm{K}}_{\mathrm{B}}=3.00×{10}^{-7} \mathrm{J}$

Therefore, from the energy conservation law, total energy will be constant.

$$\begin{gathered} E_A=E_B \\ 0+q{V}_A=K_B+q{V}_B \\ {V}_A=\frac{K_B}{q}+V_B \\ {V}_B=V_A-\frac{K_B}{q} \end{gathered}$$

Now put the values in the above equation; we get,

$$\begin{gathered} {\mathrm{V}}_{\mathrm{B}}={\mathrm{V}}_{\mathrm{A}}-\frac{{\mathrm{K}}_{\mathrm{B}}}{\mathrm{q}} \\ =+30.0\mathrm{V}-\frac{\left(3.00×{10}^{-7}\mathrm{J}\right)}{\left(-6.00×{10}^{-9}\mathrm{C}\right)} \\ =+80.0\mathrm{V} \end{gathered}$$

Therefore, the electric potential at B is 80 V.

(b)

The difference between the two points and the electric field are related as:

$$\begin{gathered} {\mathrm{V}}_{\mathrm{A}}-{\mathrm{V}}_{\mathrm{B}}={∫}_0^1\mathrm{E}.\mathrm{dl} \\ {\mathrm{V}}_{\mathrm{A}}-{\mathrm{V}}_{\mathrm{B}}=\mathrm{E}{∫}_0^1\mathrm{dl} \\ {\mathrm{V}}_{\mathrm{A}}-{\mathrm{V}}_{\mathrm{B}}=\mathrm{El} \\ \mathrm{E}=\frac{{\mathrm{V}}_{\mathrm{A}}-{\mathrm{V}}_{\mathrm{B}}}{\mathrm{l}} \end{gathered}$$

Here l is the traveled distance which is 0.500 m.

Now put the values in the equation; we get,

$$\begin{gathered} \mathbf{E}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{A}}\mathbf{-}{\mathbf{V}}_{\mathbf{B}}}{\mathbf{I}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{30}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{V}\mathbf{-}\mathbf{80}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{V}}{\mathbf{0}\mathbf{.}\mathbf{500}\mathbf{}\mathbf{m}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{100}\mathbf{}\mathbf{V}\mathbf{/}\mathbf{m} \end{gathered}$$

Therefore, the magnitude is 100 V/m, and the direction of the magnetic field is point B to point A