Question

Two identical negative charges Q are placed on equal distance r of a positive charge Qb (see photo). What is then the value of Q causing the charges (relative to each other) to be in balance? Thank you!

Answer 1

The result is `Q = 4*Q_b`.

Explanation:

Let's look at the Q on the left. It will be attracted to `Q_b` and repelled from the other Q. We need to find the value of Q, in terms of `Q_b` that will balance those 2 forces.

The equation for that state will state that the sum of those 2 forces = 0.

`(k*(-Q)*Q_b)/r^2 + (k*(-Q)*(-Q))/(2*r)^2 = 0`

`(k*(-Q)*Q_b)/r^2 + (k*Q^2)/(4*r^2) = 0`

`(k*Q*Q_b)/r^2 = (k*Q^2)/(4*r^2)`

`(cancel(k)*cancel(Q)*Q_b)/cancel(r^2) = (cancel(k)*Qcancel(^2))/(4*cancel(r^2))`

The result is `Q = 4*Q_b`.

I hope this helps,
Steve