Question

In the diagram, q1 = +6.39*10^-9 C. Theelectric field at point P is zero. What isthe value of the charge q2? Include a +or - sign.(Remember, E points away from + charges,and toward charges.)(The answer is *10^-8 C. Just fill in thenumber, not the power.)

Answer 1

As the problem tells us, the field at point P is 0, thus, the field exerted by charges q1 and q2 have the same magnitude, and exactly the opposite direction, as it can be seen on the following drawing:

Thus, we know that charge q2 will have to be a source of field (as opposed to a sink), and thus, a positive charge. Now all we have to do is find out what charge could produce a field with the same magnitude of the one from q1. As the electric field can be written as:

`E=(kq)/(d^2)`

We'll have:

`(kq_1)/(d_1^2)=(kq_2)/(d_2^2)\Rightarrow(6.39*10^(-9))/(0.424^2)=(q_2)/(0.636^2)\Rightarrow q_2=(6.39*10^(-9)*0.636^2)/(0.424^2)`
`q_2=1.43775*10^(-8)C`

Thus, our answer is q2=1.43775*10^(-8)C

Note: Your lesson requests the answer to be inserted as: +1.44