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!

enter image source here

1 Answer
Nov 3, 2017

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