Two protons are on either side of an electron as shown below:
Diagram
The electron is 30 µm away from the proton on its left and 10 µm away from the proton on its right. What is the magnitude and direction of the net electric force acting on the electron? Note that
P=1.6×10(-19) c=-e

Select one:
a. 2.0×10−18N to the right
b. 7.0×10−24N to the right
c. 2.0×10−18N to the left
d. −9.59×10−24N to the right
e. 7.0×10−24N to the

Let [tex]q_1[/tex] and [tex]q_2[/tex] denote the magnitude of electric charge on two point charges. Let [tex]r[/tex] denote the distance between these two charges.

By Coulomb's Law, the magnitude of the force between these two point charges would be:

[tex]\displaystyle F = \frac{k\, q_1\, q_2}{r^2}[/tex].

The unit of distance in the Coulomb's constant here[tex]k \approx 8.987552 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2}[/tex] is meters. However, the distance between the electron and each proton is given in micrometers, [tex]\rm \mu m[/tex]. Note that [tex]1\; \rm \mu m = 10^{-6}\; \rm m[/tex].

Convert these units to meters:

[tex]30 \; \rm \mu m = 30 \times 10^{-6}\; \rm m[/tex].

[tex]10 \; \rm \mu m = 10 \times 10^{-6}\; \rm m[/tex].

The question states that the magnitude of electric charge on each proton and each electron is [tex]1.6\times 10^{-19}\; \rm C[/tex].

Using these information, find the magnitude of the electric force between the electron and each proton:

Between the electron and the proton on the left-hand side of the electron: [tex]\begin{aligned}F &= \frac{k\, q_1\, q_2}{r^2}\\ &\approx \frac{\left(8.987552\times 10^{9}\; \rm N \cdot m^2\cdot C^{-2}\right) \times \left(1.6 \times 10^{-19}\; \rm C\right) \times \left(1.6 \times 10^{-19}\; \rm C\right)}{\left(30 \times 10^{-6}\; \rm m\right)} \\ &\approx 0.26 \times 10^{-18}\; \rm N\end{aligned}[/tex].
Between the electron and the proton on the right-hand side of the electron:
[tex]\begin{aligned}F &= \frac{k\, q_1\, q_2}{r^2}\\ &\approx \frac{\left(8.987552\times 10^{9}\; \rm N \cdot m^2\cdot C^{-2}\right) \times \left(1.6 \times 10^{-19}\; \rm C\right) \times \left(1.6 \times 10^{-19}\; \rm C\right)}{\left(10 \times 10^{-6}\; \rm m\right)} \\ &\approx 2.56 \times 10^{-18}\; \rm N\end{aligned}[/tex].

The two protons in this question will both attract the electron. Therefore, the force between the electron and the proton on the left-hand side of the electron would point to the left. Similarly, the force between the electron the proton on the right-hand side of the electron would point to the right.

The magnitude of the net electric force on the electron would be:

[tex]2.56 \times 10^{-18}\; \rm N - 0.26\times 10^{-18}\;\rm N \approx 2\times 10^{-18}\; \rm N[/tex].

These two forces act along the same line in opposite direction. Therefore, the resultant force of these two force would be in the direction of the larger force of the two.

In this question, the electric force between the electron and the proton on the right-hand side of the electron is larger. Hence, the net electric force of the protons on the electron should point to the right (towards the proton that is closer to the electron.)