Question #177920

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).


1
Expert's answer
2021-04-05T11:10:18-0400

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

(i) The electric field E1E_1directed to the right from the point pp on the xx-axis (or away from the charge q1q_1because it is positively charged). The electric field E2E_2 directed to the right from the point pp on the xx-axis (or toward the charge q2q_2 because it is negatively charged). Then, we can write the net electric field:

Enet=E1+E2=k(q1rp2+q2(rrp)2),E_{net}=E_1+E_2=k(\dfrac{|q_1|}{r_p^2}+\dfrac{|q_2|}{(r-r_p)^2}),Enet=9109 Nm2C2(2109 C(0.2 m)2+5109 C(0.8 m0.2 m)2)=575 NC.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 E1E_1directed to the right from the point pp on the xx-axis (or away from the charge q1q_1because it is positively charged). The electric field E2E_2 directed to the left from the point pp on the xx-axis (or toward the charge q2q_2 because it is negatively charged). Then, we can write the net electric field:


Enet=E1E2=k(q1rp2q2(rpr)2),E_{net}=E_1-E_2=k(\dfrac{|q_1|}{r_p^2}-\dfrac{|q_2|}{(r_p-r)^2}),Enet=9109 Nm2C2(2109 C(1.2 m)25109 C(1.2 m0.8 m)2)=269 NC.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 E1E_1directed to the left from the point pp on the xx-axis (or away from the charge q1q_1because it is positively charged). The electric field E2E_2 directed to the right from the point pp on the xx-axis (or toward the charge q2q_2 because it is negatively charged). Then, we can write the net electric field:


Enet=E1+E2=k(q1rp2+q2(r+rp)2),E_{net}=-E_1+E_2=k(-\dfrac{|q_1|}{r_p^2}+\dfrac{|q_2|}{(r+r_p)^2}),Enet=9109 Nm2C2(2109 C(0.2 m)2+5109 C(0.8 m+0.2 m)2)=405 NC.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=Enetq.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 NC1.61019 C=9.21017 C.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 NC1.61019 C=4.31017 C.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 NC1.61019 C=6.51017 C.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).


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