Answer to Question #111716 in Electric Circuits for Ashley Nofa

Question

Two charges each +4 nC 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:
x = -2 m d) x = 10 m
x = 2 m e) At what point on the x-axis is the electric field zero?
x = 6 m f) Sketch Ex versus x

Answer 1

According to the superposition principle, the net electric field from the two charges is the sum of the fields that would have been caused by each charges individually

So, let's find the individual electric fields in each point using Coulomb's law. Assume that $q = 4\cdot10^{-9} C$ is our charge and $k = 9\cdot10^9 N⋅m^2⋅C^{−2}$ is Coulomb's constant.

1. $x = -2$ . The field from the first charge: $E_1 = -k\dfrac{q}{x^2} = -9 \space V/m$ , the field from the second charge: $E_2 = -k\dfrac{q}{(8-x)^2} = -0.36 \space V/m$ . The net field is: $E = E_1 + E_2 = -9.36\space V/m$ .
2. $x = 10.$ The field from the first charge: $E_1 = k\dfrac{q}{x^2} = 0.36 \space V/m$ , the field from the second charge: $E_2 = k\dfrac{q}{(8-x)^2} = 9 \space V/m$ . The net field is: $E = E_1 + E_2 = 9.36\space V/m$ .
3. $x = 2.$ The field from the first charge: $E_1 = k\dfrac{q}{x^2} = 9 \space V/m$, the field from the second charge: $E_2 = -k\dfrac{q}{(8-x)^2} = -1 \space V/m$. The net field is: $E = E_1 + E_2 = 8\space V/m$.
4. $x = 6.$ The field from the first charge: $E_1 = k\dfrac{q}{x^2} = 1 \space V/m$, the field from the second charge: $E_2 = -k\dfrac{q}{(8-x)^2} = -9 \space V/m$. The net field is: $E = E_1 + E_2 = -8\space V/m$
5. The point on the x-axis where the electric field is zero is the poin in the middle of two charges, where both forces are compensated. Thus in should be $x = 4$.
6. The sketch of Ex versus x: