For a path starting at B and going to A (Figure 16.9), calculate (a) the change in electric potential, (b) the potential energy change for the system when a proton moves from B to A, and (c) the potential energy change for the system when an electron moves from B to A. For a path starting at B and going to C, calculate (d) the change in electric potential, (e) the potential energy change for the system when a proton moves from B to C, and (f) the potential energy change for the system when an electron moves from B to C.

The magnitude of the electric field for a charged particle (positive or negative) relies on the change in the electric potential energy of that particle.

Also, the potential energy change with the amount of distance the particle travels in the electric filed.

The value of the uniform electric field is,$\overrightarrow{E}=\left(500,0,0\right)N/C$.

The location of the point A is,$A\left(-0.4,0,0\right)m$ .

The location of the point B is, $B\left(0.2,0,0\right)m$.

The location of the point C is,$C\left(0.2,-0.3,0\right)m$ .

The travel distance from the location B to location A is given by,

$$\begin{gathered} ∆{\overrightarrow{l}}_{BA}=\left(-0.4,0,0\right)m-\left(0.2,0,0\right)m \\ ∆{\overrightarrow{l}}_{BA}=\left(-0.6,0,0\right)m \end{gathered}$$

Then the formula for the change in electric potential from B to A is given by,

$$\begin{gathered} ∆V_{BA}=-\left(E_x∆x+E_y∆y+E_z∆z\right) \\ ∆V_{BA}=-\left(500N/C\times -0.6m+0\times 0+0\times 0\right) \\ ∆V_{BA}=-\left(-300+0+0\right)N.m/C\times \frac{1V}{1N.m/C} \\ ∆V_{BA}=300V \end{gathered}$$

Hence, the change in electric potential from B to A is 300 V.

The value of the charge on a proton is,

$q_P=1.6x{10}^{-19}\mathrm{C}$

The formula for the potential energy change when a proton moves from B to A is given by,

role="math" localid="1657086281641" $$\begin{gathered} ∆U=q_p\times ∆V_{BA} \\ ∆U=1.6\times {10}^{-19}C\times 300V\times \frac{1J}{1C.V} \\ ∆U=4.8\times {10}^{-17}J \end{gathered}$$

Hence, the potential energy change when a proton moves from B to A isrole="math" localid="1657086290756" $4.8\times {10}^{-17}J$.

The value of the charge on anelectron is,

$q_e=-1.6x{10}^{-19}C$

The formula for the potential energy change when an electron moves from B to A is given by,

$$\begin{gathered} ∆U=q_e\times ∆V_{BA} \\ ∆U=1.6\times {10}^{-19}C\times 300V\times \frac{1J}{1C.V} \\ ∆U=-4.8\times {10}^{-17}J \end{gathered}$$

Hence, the potential energy change when anelectron moves from B to A is $-4.8\times {10}^{-17}J$.

The travel distance from the location B to location C is given by,

$$\begin{gathered} ∆{\overrightarrow{l}}_{BC}=\left(0.2,-0.3,0\right)m-\left(0.2,0,0\right)m \\ ∆{\overrightarrow{l}}_{BC}=\left(0,-0.3,0\right)m \end{gathered}$$

Then the formula for the change in electric potential from B to C is given by,

$$\begin{gathered} ∆V_{BC}=-\left(E_x∆x+E_y∆y+E_z∆z\right) \\ ∆V_{BC}=-\left(500N/C\times 0+0\times -0.3m-0\times 0\right) \\ ∆V_{BC}=\left(0+0+0\right)N.m/C\times \frac{1V}{1Nm/C} \\ ∆V_{BC}=0V \end{gathered}$$

Hence, the change in electric potential from B to C is 0 V.

The value of the charge on a proton is,

$q_p=1.6x{10}^{-19}\mathrm{C}$

The formula for the potential energy change when a proton moves from B to C is given by,

$$\begin{gathered} ∆U=q_p\times ∆V_{BC} \\ ∆U=1.6\times {10}^{-19}C\times 0V \\ ∆U=0 \end{gathered}$$

Hence, the potential energy change when a proton moves from B to C is 0.

The value of the charge on an electron is,

$q_e=-1.6x{10}^{-19}C$

The formula for the potential energy change when an electron moves from B to C is given by,

$$\begin{gathered} ∆U=q_e\times ∆V_{BC} \\ ∆U=1.6\times {10}^{-19}C\times 0V \\ ∆U=0 \end{gathered}$$

Hence, the potential energy change when an electron moves from B to C is 0.