Question

Two point charges, 3.4 μC and -2.0 μC , are placed 5.0 cm apart on the x axis. Assume that the negative charge is at the origin, and the positive x-axis is directed from the negative charge to the positive. At what points along the x axis is the potential zero? Let V=0 at r=∞ .

Answer 1

The electric field is zero at x = -16.45cm

Data;

• q1 = 3.4 μC
• q2 = -2.0 μC
• distance = 5cm

The Electric Field at point 0

As the 3μC is larger than -2.0μC  and the charges are opposite sign. The electric field will be zero at the negative axis.

Let the point be at x.

For an electric field to be equal to zero;

$k(\frac{q_1}{d_1})^2 + k(\frac{q_2}{d_2})^2 = 0\\\frac{3.4}{(5-x)^2} - \frac{2}{x^2} = 0$

Let's solve for x using mathematical methods.

$\frac{3.4x^2 - 2(5-x)^2}{(5-x)^2x^2}= 0\\ 3.4x^2 - 2(5-x^2) = 0\\3.4x^2 - 50 - 2x^2 + 20x = 0\\1.4x^2 +20x - 50 = 0$

Solving the above quadratic equation;

$x = -16.45cm$

The electric field is zero at x = -16.45cm

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