Question

What can you say about two charges \({q_1}\) and \({q_2}\), if the electric field one-fourth of the way from \({q_1}\) to \({q_2}\) is zero?

Short answer

The charge \({q_2}\) is \(9\) times larger than \({q_1}\).

Step-by-step solution

Electric field

Electric field is defined as the force experienced by a unit positive charge when place in the region of another charge. The expression for the electric field is,

\(E = \frac{{Kq}}{{{r^2}}}\)

Here,\(K\)is the electrostatic force constant,\(q\)is the charge and\(r\)is the distance of point of consideration from the charge.

For a system of multiple charge, the electric field at a particular point is the vector sum of all the electric field at that point.

Electric field one-fourth of the way

The electric field one-fourth of the way from \({q_1}\) to \({q_2}\) is represented as,

Electric field one-fourth of the way from\({q_1}\)to\({q_2}\)

Here,\({E_1}\)is the electric field due to charge\({q_1}\), and\({E_2}\)is the electric field due to charge\({q_2}\).

The electric field due to charge\({q_1}\)is,

\({E_1} = \frac{{K{q_1}}}{{{x^2}}}\)

The electric field due to charge\({q_2}\)is,

\({E_2} = \frac{{K{q_2}}}{{{{\left( {3x} \right)}^2}}}\)

The net electric field one-fourth of the way from\({q_1}\)to\({q_2}\)is,

\(\begin{array}{c}E = {E_1} - {E_2}\\ = \frac{{K{q_1}}}{{{x^2}}} - \frac{{K{q_2}}}{{{{\left( {3x} \right)}^2}}}\end{array}\)

Since the electric field one-fourth of the way from\({q_1}\)to\({q_2}\)is zero i.e.,\(E = 0\). Therefore,

\(\begin{array}{c}0 = \frac{{K{q_1}}}{{{x^2}}} - \frac{{K{q_2}}}{{{{\left( {3x} \right)}^2}}}\\\frac{{K{q_1}}}{{{x^2}}} = \frac{{K{q_2}}}{{{{\left( {3x} \right)}^2}}}\\{q_2} = \frac{{{{\left( {3x} \right)}^2}}}{{{x^2}}}{q_1}\\{q_2} = 9{q_1}\end{array}\)

Hence, the charge\({q_2}\)is\(9\)times larger than\({q_1}\).