Question

Locations A, B and C are in a region of uniform electric field, as shown in Figure 16.66. Location A is at (-0.3,0,0)m. Location B is at (0.4,0,0)m. In the region the electric field has the value (850,400,0)N/C . For a path starting at A and ending at B calculate:

(a) the displacement vector $△\stackrel{-}{I}$.

(b) the change in electric potential,

(c) the potential energy change for the system when a proton moves from A to B ,

(d) the potential energy change for the system when an electron moves from A to B.

Short answer

a. The displacement vector is, $136\left({10}^{-19}\right)$.

b. the change in electric potential is, 85 volts.

c. the potential energy change for the system when a proton moves from A to B is decreases.

d. the potential energy change for the system when an electron moves from A to B is, increase.

Step-by-step solution

Given Information.

Locations A, B and C are in a region of uniform electric field,

$\begin{array}{l}\overrightarrow{V_A}=-03m\\ {\overrightarrow{V}}_B=0.4m\end{array}$

The electric field has the value,$\overrightarrow{E}=-850400NlC$

Concept used in the question.

The displacement vector from one point to another is an arrow with its tail at the first point and its tip at the second. The magnitude (or length) of the displacement vector is the distance between the points and is represented by the length of the arrow.

Displacement vector.

a).

The displacement vector is,

$\begin{array}{l}W=-\left(q\overrightarrow{E}.\overrightarrow{S}\right)\\ \overrightarrow{S}=\overrightarrow{V_B}+\overrightarrow{V_A}\\ \overrightarrow{S}=\left(-0.3\right)\overrightarrow{i}+\left(0.4\right)\overrightarrow{i}\\ \overrightarrow{S}=\left(0.1\right)\overrightarrow{i}\end{array}$

$$\begin{gathered} △u=\left(-1.6\left({10}^{-19}\right)\right)\left(-\left(850\overrightarrow{i}\right)+\left(400\overrightarrow{j}\right)\right) \\ △u=\left(1.60\right)\left(850\overrightarrow{i}\right)\left(0.1\right)\left({10}^{-19}\right) \\ △u=136\left({10}^{-19}\right) \end{gathered}$$

The change in electric potential.

b).

$$\begin{gathered} △u=\left(-1.6\left({10}^{-19}\right)\right)\left(-\left(850\overrightarrow{i}\right)+400\overrightarrow{j}\right) \\ V=\overrightarrow{-E}.\overrightarrow{r} \\ V=-\left(-\left(850\overrightarrow{i}\right)+400\overrightarrow{j}\right)\left(-\left(0.1\overrightarrow{i}\right)\right) \\ V=\left(850\overrightarrow{i}\right)\left(-\left(0.1\overrightarrow{i}\right)\right) \\ V=85Volts \end{gathered}$$

The potential energy change for the system when a proton moves from A to B.

c).

As proton moves from A to B its potential energy converts into kinetic energy,

So, potential energy of the system will decreases.

Hence, The potential energy change for the system when a proton moves from one point to another then it always decreases.

The potential energy change for the system when a electron moves from A to B.

d).

Asproton moves from A to B itspotential energy converts into kinetic energy,

So, potential energy of the system will increase.

when an electron moves from a region of high electric potential to a region of lower electric potential, its potential energy increases. This is because it has a negative charge and a decrease in electrical potential thus results in an increase in potential energy.

Hence, The potential energy change for the system when a proton moves from one point to another then it always increases.