# 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?

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
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