Question

Two charges, each 9 µC, are on the x axis, one at the origin and the other at x = 8 m. Find the electric field on the x axis at each of the following locations.
(a) x =-2 m N/C i
(b) x = 2 m
(c) x 6 m N/C
(d) x = 10 m N/C
(e) At what point on the x axis is the electric field zero?

Answer 1

a) Electric field at x = -2 m: 21,060 N/C to the left

b) Electric field at x = 2 m: 18,000 N/C to the right

c) Electric field at x = 6 m: 18,000 N/C to the left

d) Electric field at x = 10 m: 21,060 N/C to the right

e) Electric field is zero at x = 4 m

Explanation:

a)

The electric field produced by a single-point charge is given by

$E=k\frac{q}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

q is the magnitude of the charge

r is the distance from the charge

Here we have two charges of

$q=9\mu C = 9\cdot 10^{-6} C$

each. Therefore, the net electric field at any point in the space will be given by the vector sum of the two electric fields. The two charges are both positive, so the electric field points outward of the charge.

We call the charge at x = 0 as $q_0$ , and the charge at x = 8 m as $q_8$.

For a point located at x = -2 m, both the fields $E_0$ and $E_8$ produced by the two charges point to the left, so the net field is the sum of the two fields in the negative direction:

$E=-\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=-kq(\frac{1}{(-2)^2}+\frac{1}{(8-(-2))^2})=-21060 N/C$

b)

In this case, we are analyzing a point located at

x = 2 m

The field produced by the charge at x = 0 here points to the right, while the field produced by the charge at x = 8 m here points to the left. Therefore, the net field is given by the difference between the two fields, so:

$E=\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(2)^2}-\frac{1}{(8-2)^2})=18000 N/C$

And since the sign is positive, the direction is to the right.

c)

In this case, we are considering a point located at

x = 6 m

The field produced by the charge at x = 0 here points to the right again, while the field produced by the charge at x = 8 m here points to the left. Therefore, the net field is given by the difference between the two fields, as before; so:

$E=\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(6)^2}-\frac{1}{(8-6)^2})=-18000 N/C$

And the negative sign indicates that the electric field in this case is towards the left.

d)

In this case, we are considering a point located at

x = 10 m

This point is located to the right of both charges: therefore, the field produced by the charge at x = 0 here points to the right, and the field produced by the charge at x = 8 m here points to the right as well. Therefore, the net field is given by the sum of the two fields:

$E=\frac{kq_0}{(0-x)^2}+\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(10)^2}+\frac{1}{(8-(10))^2})=21060 N/C$

And the positive sign means the field is to the right.

e)

We want to find the point with coordinate x such that the electric field at that location is zero. This point must be in between x = 0 and x = 8, because that is the only region where the two fields have opposite directions. Therefore, te net field must be

$E=\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(-x)^2}-\frac{1}{(8-x)^2})=0$

This means that we have to solve the equation

$\frac{1}{x^2}-\frac{1}{(8-x)^2}=0$

Re-arranging it,

$\frac{1}{x^2}-\frac{1}{(8-x)^2}=0\\\frac{(8-x)^2-x^2}{x^2(8-x)^2}=0$

So

$(8-x)^2-x^2=0\\64+x^2-16x-x^2=0\\64-16x=0\\64=16x\\x=4 m$

So, the electric field is zero at x = 4 m, exactly halfway between the two charges (which is reasonable, because the two charges have same magnitude)

Learn more about electric fields:

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Answer 2

Answer:

(a) E=-21,060 N/C

(b) E=18,000 N/C

(c) E=-18,000 N/C

(d) E= 22,500 N/C

(e) x=4 m

Explanation:

Electric Field

The electric field generated by a point charge Q at a distance d is given by

$\displaystyle E=\frac{k\ q}{r^2}$

where k is the Coulomb's constant, q is the value of the charge and r is the distance.

The electric field is a vector and its direction is pushing away from the positive charge and pulling to the negative charge.

Please refer to the image below to better find directions and magnitudes of the calculations

(a) At x=-2, both charges push to the left, making the total electric field be negative and additive

Charge 1 is at 2 m

$\displaystyle E_1=-\frac{9\times 10^9\ 9\times 10^{-6}}{2^2}=-20,250\ N/c$

Charge 2 is at 10 m

$\displaystyle E_2=-\frac{9\times 10^9\ 9\times 10^{-6}}{10^2}=-810\ N/c$

Total field:

$E=-20,250-810=-21,060\ N/c$

(b) At x=2 m, q1 pushes to the right and q2 pushes to the left.

$\displaystyle E_1=\frac{9\times 10^9\ 9\times 10^{-6}}{2^2}=20,250\ N/c$

$\displaystyle E_2=-\frac{9\times 10^9\ 9\times 10^{-6}}{6^2}=-2,250\ N/c$

$E=E_1+E_2$

$E=18,000\ N/c$

(c) At x=6 m, q1 pushes to the right and q2 pushes to the left.

$\displaystyle E_1=\frac{9\times 10^9\ 9\times 10^{-6}}{6^2}=2,250\ N/c$

$\displaystyle E_2=\frac{9\times 10^9\ 9\times 10^{-6}}{2^2}=-20,250\ N/c$

$E=2,250\ N/c-20,250\ N/c$

$E=-18,000\ N/c$

(d) At x=10 m, both charges push to the right

$\displaystyle E_1=\frac{9\times 10^9\ 9\times 10^{-6}}{10^2}=810\ N/c$

$\displaystyle E_2=\frac{9\times 10^9\ 9\times 10^{-6}}{2^2}=20,250\ N/c$

$E=2,250\ N/c+20,250\ N/c$

$E=22,500\ N/c$

(e) We must find a point where

$E_1=E_2$

pointing in opposite directions. Let's call x the distance from the origin where that happens, then

$\displaystyle \frac{k\ q}{x^2}=\frac{k\ q}{(8-x)^2}$

Simplifying

$x^2=(8-x)^2$

It has only one solution

$x=4$

So at x=4 both fields are equal and opposite, so the total field is zero