Question

A negative point charge ${\mathbf{q}}_{\mathbf{1}}$= -4.00 nC is on the x-axis at x = 0.60 m. A second point charge ${\mathbf{q}}_{\mathbf{2}}$is on the x-axis at x = -1.20 m. What must the sign and magnitude ${\mathbf{q}}_{\mathbf{2}}$be for the net electric field at the origin to be

(a) 50.0 N/C in the +x direction and

(b) 50.0 N/C in the –x-direction?

Short answer

a) $q_2$is -8.0nC when the net electric field at the origin is 50.0 N/C in the +x direction.

b) $q_2$is -24.0nC when the net electric field at the origin is 50.0 N/C in the +x direction

Step-by-step solution

Calculation of charge in +x direction.

Given ${\mathrm{r}}_1=0.6\mathrm{m},{\mathrm{r}}_2=1.2\mathrm{m}$and

$q=-4nc$

a) The net electric field at the origin of the first charge is positive x is equal to

${\mathrm{E}}_1=\frac{{\mathrm{kq}}_1}{{{\mathrm{r}}^2}_1}=\frac{9\times 10^9\times 4\times {10}^{-9}}{{(0.6)}^2}=100\mathrm{N}/\mathrm{C}$ (1)

so the electric field due to the second charge must be -x and its magnitude must be 50N/C so the second charge is negative and is equal to

${\mathrm{E}}_1=\frac{{\mathrm{kq}}_1}{{{\mathrm{r}}^2}_1}=q=\frac{50\times {\left(1.2\right)}^2}{9\times {10}^9}=8.0\times {10}^{-9}c8.0nC$

So the charge is $q_2$=-8nC

Calculation of charge in -x direction.

The net electric field at the origin of the first charge is positive x and is equal to

${\mathrm{E}}_1=\frac{{\mathrm{kq}}_1}{{{\mathrm{r}}^2}_1}=\frac{9\times 10^9\times 4\times {10}^{-9}}{{\left(0.6\right)}^2}=100\mathrm{N}/\mathrm{C}$ (1)

so the electric field due to the second charge must be -x and its magnitude must be 150N/C so the second charge is negative and its magnitude is given by

$\mathrm{q}=\frac{50\times {\left(1.2\right)}^2}{9\times {10}^9}=8.0\times {10}^{-9}\mathrm{c}8.0\mathrm{nC}$

So the charge is $q_2$= - 24nC