# Summing two electric fields

1. Oct 13, 2011

### nathangrand

The electrical system of typical thundercloud can be represented by a vertical dipole consisting of a charge of +40C at a height of 10km and a charge of -40C vertically below it at a height of 6km. What is the electric field at the ground directly beneath the thundercloud.

I am simply adding the the field from each point charge at the ground together but get the wrong answer...any suggestions?

Eground= 40/4*∏*ε0(1/10000^2 - 1/6000^2)

2. Oct 13, 2011

### SammyS

Staff Emeritus
At the point on the ground, where you are calculating the E field:
What is the direction of the E field due to a positive charge which is above?

What is the direction of the E field due to a negative charge which is above?

3. Oct 13, 2011

### nathangrand

+ve into ground, -ve charge field up?

4. Oct 13, 2011

### SammyS

Staff Emeritus
Yes. Now look at the signs you have in your calculation.

5. Oct 13, 2011

### nathangrand

A sign error isn't my problem - it's the magnitude I get wrong...

6. Oct 13, 2011

### SammyS

Staff Emeritus
Place proper parentheses so the π and ε0 are in the denominator.