91Ó°ÊÓ

We are given a charge q = -3.00 nC on the x-axis at r₁ = 1.20 m from the origin. A second point charge, Q, is on the - x-axis at distance r = 0.600 m.

(a)The electric field at the origin is due to the net electric field from the two charges q and Q. Let us calculate the electric field by the charge q then calculate the electric field by the charge Q to find its charge magnitude.

The electric field at origin produced by q is

$\begin{array}{r}{E}_q=k\frac{q}{r_1^2}\\ =\left(9×{10}^{9}N.m^2/C^2\right)\frac{\left|−3.00×{10}^{−9}\mathrm{C}\right|}{(1.20\mathrm{m}{)}^2}\\ =18.75\mathrm{N}/\mathrm{C}\end{array}$

As the direction of the net electric field is in +x direction, this means the electric field E at the origin

$\begin{array}{r}E=E_q+E_Q\\ E_Q=E−E_q\\ E_Q=45.0\mathrm{N}/\mathrm{C}−18.75\mathrm{N}/\mathrm{C}\\ E_Q=26.25\mathrm{N}\end{array}$

The E is away from Q this means the charge Q is, a positive charge, and the magnitude of Q could be calculated

$\begin{array}{r}{E}_Q=k\frac{\left|Q\right|}{r_2^2}\\ Q=\frac{E_Qâ‹\dots r_2^2}{k}\\ Q=\frac{1}{k}\left(E_Qâ‹\dots r_2^2\right)\end{array}$

Now substitute the given values in the above equation

$\begin{array}{r}=\frac{1}{\left(9×{10}^{9}N{m}^2/C^2\right)}\left((26.25\mathrm{N}/\mathrm{C})(0.600\mathrm{m}{)}^2\right)\\ =1.05×{10}^{−9}\mathrm{C}\\ =1.05n\mathrm{C}\end{array}$

Hence the magnitude of the charge Q is 1.05nC and it is positive.

(b) When the net electric field at the origin is in the negative direction of x, this means the magnitude of E is less than the magnitude of the electric field produced due to charge Q. where the two electric fields of Q and q are in opposite direction.

And the electric field of Q will be

$\begin{array}{r}{E}_Q=E+E_q\\ =45.0\mathrm{N}/\mathrm{C}+18.75\mathrm{N}/\mathrm{C}\\ =63.75\mathrm{N}/\mathrm{C}\end{array}$

As the net electric field is toward Q, then the charge Q is negative

Now substitute the given values to the equation (1),

$\begin{array}{r}Q=\frac{1}{k}\left(E_Qâ‹\dots r_2^2\right)\\ =\frac{1}{\left(9×{10}^{9}N.m^2/C^2\right)}\left((63.75\mathrm{N}/\mathrm{C})(0.600\mathrm{m}{)}^2\right)\\ =2.55×{10}^{−9}\mathrm{C}\\ =2.55\mathrm{nC}\end{array}$

Hence the magnitude of the charge Q is 2.55 nC and it is negative.