Question

Find the force on a negative charge that is placed midway between two equal positive charges. All charges have the same magnitude.

Answer 1

Answer: The force on a negative charge that is placed midway between two equal positive charges is zero when all charges have the same magnitude.

Explanation:

Let us assume that

$q_{1} = q_{2} = +q$

$q_{3} = -q$

As $q_{3}$ is the negative charge and placed midway between two equal positive charges ($q_{1}$ and $q_{2}$).

Total distance between $q_{1}$ and $q_{2}$ is 2r. This means that the distance between $q_{1}$ and $q_{3}$, $q_{2}$ and $q_{3}$ = d = r

Now, force action on charge $q_{3}$ due to $q_{1}$ is as follows.

$F_{31} = k(\frac{q_{1} \times q_{3}}{d^{2}})$

where,

k = electrostatic constant = $9 \times 10^{9} Nm^{2}/C^{2}$

Substitute the values into above formula as follows.

$F_{31} = k(\frac{q_{1} \times q_{3}}{d^{2}})\\= 9 \times 10^{9} (\frac{q \times (-q)}{r^{2}})\\= - 9 \times 10^{9} (\frac{q^{2}}{r^{2}})$ ... (1)

Similarly, force acting on $q_{3}$ due to $q_{1}$ is as follows.

$F_{32} = k \frac{q_{2}q_{3}}{d^{2}}\\= -9 \times 10^{9} \frac{q^{2}}{r^{2}}$   ... (2)

As both the forces represented in equation (1) and (2) are same and equal in magnitude. This means that the net force acting on charge $q_{3}$ is zero.

Thus, we can conclude that the force on a negative charge that is placed midway between two equal positive charges is zero when all charges have the same magnitude.