Summing two electric fields

[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)

Answer should be 12.8KV/m upwards

 

[SammyS]

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?
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[nathangrand]

+ve into ground, -ve charge field up?

 

[SammyS]

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

 

[nathangrand]

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

 

[SammyS]

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

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