A charge of \(-1 \mathrm{nC}\) is located at the origin in free space. What charge must be located at \((2,0,0)\) to cause \(E_{x}\) to be zero at \((3,1,1)\) ?

We are given a charge of -1nC at the origin in free space. We need to find the charge at point (2,0,0) such that E_x at the point (3,1,1) becomes zero. Let the charge at point (2,0,0) be \(q_2\).

First, calculate the distance between each charge and the point (3,1,1). For the charge at the origin: \(r_1 = \sqrt{(3-0)^2+(1-0)^2+(1-0)^2}=\sqrt{9+1+1}= \sqrt{11}\) For the charge at point (2,0,0): \(r_2=\sqrt{(3-2)^2+(1-0)^2+(1-0)^2}=\sqrt{1+1+1}= \sqrt{3}\)

Now, we need to find the electric field components at point (3,1,1) due to both charges. For the charge at the origin: \(E_{1x} = \frac{kq_1 (x_1-x)}{(r_1)^3} = \frac{k(-1 \times 10^{-9})(3-0)}{(\sqrt{11})^3}\) \(E_{1y} = \frac{kq_1 (y_1-y)}{(r_1)^3} = \frac{k(-1 \times 10^{-9})(1-0)}{(\sqrt{11})^3}\) \(E_{1z} = \frac{kq_1 (z_1-z)}{(r_1)^3} = \frac{k(-1 \times 10^{-9})(1-0)}{(\sqrt{11})^3}\) For the charge at point (2,0,0), let's not calculate the actual values of \(E_{2x}\), \(E_{2y}\) and \(E_{2z}\), but set up the expressions for them as they depend on q2, which we are trying to find. \(E_{2x} = \frac{kq_2 (x_2-x)}{(r_2)^3} = \frac{kq_2 (3-2)}{(\sqrt{3})^3}\) \(E_{2y} = \frac{kq_2 (y_2-y)}{(r_2)^3} = \frac{kq_2 (1-0)}{(\sqrt{3})^3}\) \(E_{2z} = \frac{kq_2 (z_2-z)}{(r_2)^3} = \frac{kq_2 (1-0)}{(\sqrt{3})^3}\)

Since the x-component of the total electric field at point (3,1,1) needs to be zero, we can write: \(E_{1x}+E_{2x}=0\) \(\frac{k(-1 \times 10^{-9})(3-0)}{(\sqrt{11})^3}+\frac{kq_2 (3-2)}{(\sqrt{3})^3}=0\) Now, simplify the equation and solve for q2: \(q_2=\frac{-1 \times 10^{-9}(\sqrt{3})^3}{(\sqrt{11})^3}\) \(q_2 \approx -0.61 \times 10^{-9} C\)

To cause E_x to be zero at point (3,1,1), the charge q2 must be located at (2,0,0) with an approximate charge of \(-0.61 \mathrm{nC}\).