# Calculating charge when given electric field

## Homework Statement

Charges of 5 µC are located at x = 0, y = 2.0 m and at x = 0, y = -2.0 m. Charges Q are located at x = 4.0 m, y = 2.0 m and at x = 4.0 m, y = -2.0 m. The electric field at x = 0, y = 0 is (9e3 N/C) . Determine Q.

E = kQ/r^2

## The Attempt at a Solution

I converted my units to C so I could use the k constant of 8.99e9 N*m^2/C^2
I know that all the vertical components get cancelled out so the only force is the the x positive direction.

I set up my equation like this E= kQ/r^2 with E being 9e3 and r^2 being 20

I end up with 9e3=kQ/10 and when I solve for Q I get 1e-5 C and when I convert my answer to microcoloumbs I get 10.

I think I am having problems with my conversions. Any help would be greatly appreciated!!

SammyS
Staff Emeritus
Homework Helper
Gold Member

## Homework Statement

Charges of 5 µC are located at x = 0, y = 2.0 m and at x = 0, y = -2.0 m. Charges Q are located at x = 4.0 m, y = 2.0 m and at x = 4.0 m, y = -2.0 m. The electric field at x = 0, y = 0 is (9e3 N/C) . Determine Q.

E = kQ/r^2

## The Attempt at a Solution

I converted my units to C so I could use the k constant of 8.99e9 N*m^2/C^2
I know that all the vertical components get cancelled out so the only force is the the x positive direction.

I set up my equation like this E= kQ/r^2 with E being 9e3 and r^2 being 20

I end up with 9e3=kQ/10 and when I solve for Q I get 1e-5 C and when I convert my answer to microcoloumbs I get 10.

I think I am having problems with my conversions. Any help would be greatly appreciated!!
Hello jakethelocker. Welcome to PF !

Don't forget that E is a vector quantity. Look at the x and the y components of the field due to the Qs.

I got it! Thank you for your help :)

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