Answer to Question #177920 in Field Theory for Jo-an Balicao

Question

A +2.00 nC point charge is at the origin, and a second -5.00 nC point charge is on the x-axis at x= 0.800 m.

a) Find the electric field (magnitude and direction) at each of the following points on the x-axis: i) x= 0.200 m; ii) x= 1.20 m; iii) x= -0.200 m.

b) Find the net electric force that the two charges would exert on the electron placed at each point in part (a).

Answer 1

(a) Let's find the electric field at each of the following points on the $x$-axis. Let's choose the rightwards as the positive direction. Let denote the point on the $x$-axis as $p$.

(i) The electric field $E_1$directed to the right from the point $p$ on the $x$-axis (or away from the charge $q_1$because it is positively charged). The electric field $E_2$ directed to the right from the point $p$ on the $x$-axis (or toward the charge $q_2$ because it is negatively charged). Then, we can write the net electric field:

$$E_{net}=E_1+E_2=k(\dfrac{|q_1|}{r_p^2}+\dfrac{|q_2|}{(r-r_p)^2}),$$ $$E_{net}=9\cdot10^9\ \dfrac{Nm^2}{C^2}\cdot(\dfrac{|2\cdot10^{-9}\ C|}{(0.2\ m)^2}+\dfrac{|-5\cdot10^{-9}\ C|}{(0.8\ m-0.2\ m)^2})=575\ \dfrac{N}{C}.$$

The electric field directed rightward.

(ii) The electric field $E_1$directed to the right from the point $p$ on the $x$-axis (or away from the charge $q_1$because it is positively charged). The electric field $E_2$ directed to the left from the point $p$ on the $x$-axis (or toward the charge $q_2$ because it is negatively charged). Then, we can write the net electric field:

$$E_{net}=E_1-E_2=k(\dfrac{|q_1|}{r_p^2}-\dfrac{|q_2|}{(r_p-r)^2}),$$ $$E_{net}=9\cdot10^9\ \dfrac{Nm^2}{C^2}\cdot(\dfrac{|2\cdot10^{-9}\ C|}{(1.2\ m)^2}-\dfrac{|-5\cdot10^{-9}\ C|}{(1.2\ m-0.8\ m)^2})=-269\ \dfrac{N}{C}.$$

The sign minus means that the electric field directed leftward.

(iii) The electric field $E_1$directed to the left from the point $p$ on the $x$-axis (or away from the charge $q_1$because it is positively charged). The electric field $E_2$ directed to the right from the point $p$ on the $x$-axis (or toward the charge $q_2$ because it is negatively charged). Then, we can write the net electric field:

$$E_{net}=-E_1+E_2=k(-\dfrac{|q_1|}{r_p^2}+\dfrac{|q_2|}{(r+r_p)^2}),$$ $$E_{net}=9\cdot10^9\ \dfrac{Nm^2}{C^2}\cdot(-\dfrac{|2\cdot10^{-9}\ C|}{(0.2\ m)^2}+\dfrac{|-5\cdot10^{-9}\ C|}{(0.8\ m+0.2\ m)^2})=-405\ \dfrac{N}{C}.$$

The sign minus means that the electric field directed leftward.

(b) We can find the magnitude of the electric force from the formula:

$$F=E_{net}q.$$

Let's find the magnitude and direction of the net electric force that the two charges would exert on the electron placed at each point in part (a).

(i)

$$F=575\ \dfrac{N}{C}\cdot1.6\cdot10^{-19}\ C=9.2\cdot10^{-17}\ C.$$

As we can see from part (a), the net electric field strength directed toward the right. As we know, if a negative charge (an electron has a negative charge) is placed at the electric field, the electric force will be directed in the opposite direction to the electric field (toward the left).

(ii)

$$F=269\ \dfrac{N}{C}\cdot1.6\cdot10^{-19}\ C=4.3\cdot10^{-17}\ C.$$

As we can see from part (a), the net electric field strength directed toward the left. As we know, if a negative charge is placed at the electric field, the electric force will be directed in the opposite direction to the electric field (toward the right).

(iii)

$$F=405\ \dfrac{N}{C}\cdot1.6\cdot10^{-19}\ C=6.5\cdot10^{-17}\ C.$$

As we can see from part (a), the net electric field strength directed toward the left. As we know, if a negative charge is placed at the electric field, the electric force will be directed in the opposite direction to the electric field (toward the right).