Question

In an electric field of strength 0.50 N/C, a test charge experiences a force of 2.50 × 10^-4 N in the same direction as the field. What is the magnitude of the test charge?

Answer 1

I know the equation is Er^2/k, but I don't know what " r " is?

Answer 2

r is the distance from the test charge to the field source.

Answer 3

@PsiSquared How do we find that since it's not given?

Answer 4

The choices are:
a. -5.00 x 10^-4 C
b. -2.50 x 10^-4 C
c. +2.50 x 10^-4 C
d. +5.00 x 10^-4 C

But don't give me answer, I want to find it on my own.

Answer 5

You don't need it. Follow this:\[E=\frac{ F }{ q }\]Therefore:\[q=\frac{ F }{ E }\]If you want to see how to get there, simplify the following expression:\[q=\frac{ F }{E }=\frac{ \frac{ kq _{1}q _{2} }{ r ^{2} } }{ \frac{ kq }{ r ^{2} } }\]

Answer 6

Note that direction of the field, in this case, is indicated by the sign of the electric field.

Answer 7

Isn't E = f/q the formula for the strength of the field?

Answer 8

Yes, it is.

Answer 9

Then it would be 2.5 x 10^-4 / .50 = 0.0005 C which is D, right?

Answer 10

You got it.

Answer 11

Thank you! :D

Answer 12

No problem at all.