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.

Step-by-step 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((q_(1) * q_(3))/(d^(2)))`

where,

k = electrostatic constant =
`9 * 10^(9) Nm^(2)/C^(2)`

Substitute the values into above formula as follows.

`F_(31) = k((q_(1) * q_(3))/(d^(2)))\\= 9 * 10^(9) ((q * (-q))/(r^(2)))\\= - 9 * 10^(9) ((q^(2))/(r^(2)))` ... (1)

Similarly, force acting on
`q_(3)` due to
`q_(1)` is as follows.

`F_(32) = k (q_(2)q_(3))/(d^(2))\\= -9 * 10^(9) (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.