# Don't know where to start here :-\

NIT14
How can I solve this problem?...

An electron is placed at point P in the electric field set up by a source charge, Q. Point P is located 50 cm from Q and has an electric field strength of 1.08x10^5 N/C directed away from Q. What is the magnitude of charge Q?

## Answers and Replies

Coulombs law:

$$E(r) = \frac{Q}{4 \pi \epsilon r^2}$$

gives the electric field strength E(r) at a distance r from a point charge Q. The electrical permittivity $$\epsilon$$ can be found in a tablebook. For vacuum or air it is approx. 8.85e-12 F/m.

Last edited:
Chi Meson
Science Advisor
Homework Helper
Just a comment:
the fraction$$\frac{1}{4 \pi \epsilon}$$
is equal to the coulomb constant "k" (8.99 E9)

Originally posted by Chi Meson
Just a comment:
the fraction$$\frac{1}{4 \pi \epsilon}$$
is equal to the coulomb constant "k" (8.99 E9)
Only when:

$$\epsilon = \epsilon_0$$

If you are in a different medium there is a relative permeability $$\epsilon_r$$ in which case:

$$\epsilon = \epsilon_0\epsilon_r$$

And K is different.

Chi Meson
Science Advisor
Homework Helper
Agreed.

It was my assumption that the person asking the question was not yet at that level, and might have been taken aback by the use of epsilon when the textbook uses "k."