# Electric fields caused by multiple charges

1. Jul 1, 2009

### mcassi17

A particle with charge +7.88 μC is placed at the fixed position x = 3.00 m in an electric field of uniform strength 300 N/C, directed in the positive x direction. Find the position on the x axis where the electric field strength of the resulting configuration is zero.

the equation i have is E = kq/r^2

I have no idea where to start

2. Jul 1, 2009

### diazona

There are two electric fields in this problem: the one produced by the charge (that's what your equation is for), and the uniform one. To get the total electric field, you just add the two up (remember that they're vectors). So can you write an equation that you can solve to find the position where the total electric field is zero?

3. Jul 1, 2009

### mcassi17

So i would set it up to be -300 but that can't be cuz you can't have a negative square root.

4. Jul 1, 2009

### rl.bhat

How did you get -300?
Net electric field will be zero on the negative x-axis. If d is the distance along the negative x-axis, where electric field is zero, then E = kq/d^2. Find the magnitude of d. Subtract 3 m to find the position on negative x-axis where E is zero.