Question

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 \((3,1,1)\) ?

Short answer

The required charge at point (2,0,0) is approximately -0.61 nC.

Step-by-step solution

Understand the problem and given information

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 \(q_2\).

Calculate distance between each charge and point (3,1,1)

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): \(r_2=\sqrt{(3-2)^2+(1-0)^2+(1-0)^2}=\sqrt{1+1+1}= \sqrt{3}\)

Calculate the electric field components due to each charge

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. \(E_{2x} = \frac{kq_2 (x_2-x)}{(r_2)^3} = \frac{kq_2 (3-2)}{(\sqrt{3})^3}\) \(E_{2y} = \frac{kq_2 (y_2-y)}{(r_2)^3} = \frac{kq_2 (1-0)}{(\sqrt{3})^3}\) \(E_{2z} = \frac{kq_2 (z_2-z)}{(r_2)^3} = \frac{kq_2 (1-0)}{(\sqrt{3})^3}\)

Set up the equation for the x-component of the total electric field to be zero and solve for q2.

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: \(q_2=\frac{-1 \times 10^{-9}(\sqrt{3})^3}{(\sqrt{11})^3}\) \(q_2 \approx -0.61 \times 10^{-9} C\)

Final answer

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 \(-0.61 \mathrm{nC}\).