Question

Suppose a uniform electric field of 4N/C is in the positive x direction. When a charge is placed and at a fixed to the origin, the resulting electric field on the x axis at x =2m becomes zero. What is the magnitude of the electric field at x =4m on the x axis at this time?

Answer 1

Answer:

3N/C

Explanation:

Thinking process:

Let, electric field of 4N/C is in the positive x direction, that is + 4 NC

And the the resulting electric field on the x axis at x =2m

Therefore, based on the data above, the net charge must be in the positive direction. Thus, the magnitude is given by the equation:

$q = \frac{Er^{2} }{k} \\ = \frac{4 (4)}{9*10^{4} } \\ = \frac{16}{9}NC$

Now considering that at x = + 4 NC, the net charge will be:

$E_{net} = E_{0} + E_{q} \\ = 4 -(\frac{9*10^{} }{16}*\frac{16}{9} * 10^{-90})\\ = 3 (N/C)$