Question

A point particle with charge q = 4.9 μCis placed on the x axis at x = −15 cm.A second particle of charge Q is now placed on the x axis at x = +24 cm,and it is found that the electric field at the origin is zero. Find Q

Answer 1

Calculate the E-field for the charge you are given at point x=0. Then use that field strength to find what the other charge should be.

\[E = \frac{ kq }{ r^{2} }=\frac{ (9x10^{9})(4.9x10^{-6}) }{ (15x10^{-2})^{2} }=1.96x10^{6}\]The above is the field strength at the point x=0 due to the charge q. Now we will use this value to find the charge on the other side of the point x=0.
\[Q=\frac{ Er^{2} }{ k }=\frac{ (1.96x10^{6})(24x10^{-2})^{2} }{ (9x10^{9}) }=1.25x10^{-5}\]This has given us the magnitude of charge Q = 12.5 uC

We can ask if this makes sense. We see that charge Q is farther away from x=0 than charge q. Therefore, Q would need to be larger, and it is. Additionally, since charge q is positive, it will repel a test charge at x=0. Therefore, Q needs to repel in the opposite direction, and so it should also be positive. This is due to the fact that positive charges have radially outward E-fields.