Question

A charge of +2 C is at the origin. When charge Q is placed at 2 m along the positive x axis, the electric field at 2 m along the negative x axis becomes zero. What is the value of Q?

Answer 1

Answer:

The value of Q is - 8 C

Explanation:

Given;

the magnitude of first charge = 2 C

position of the first charge = 0

         _ (2m)                            0(+2C)                         +(2m)(Q)

------------------------------------------------------------------------------------

                                                E₁ --------------------------------->                                    

            <------------------------------------------------------------------ E₂

E₁ and E₂ are equal in magnitude but opposite in direction

| E₂ | = | E₁ |

$\frac{K*Q}{r^2} = \frac{K*2}{r^2} \\\\\frac{Q}{(4)^2} = \frac{2}{(2)^2}\\\\\frac{Q}{16} =\frac{2}{4}\\\\\frac{Q}{16} =\frac{1}{2}\\\\2Q = 16\\\\Q = 8 \ C$

Thus, since E₂ is opposite in direction to E₁, the Q = - 8 C