Homework Help: Field strength

1. Feb 9, 2008

jalen

1. The problem statement, all variables and given/known data

q=4.0x10^-6C, 8.0x10^-6C
d=2m

2. Relevant equations

Two charges of +4.0x10^-6 C and +8.0x10^-6C are placed 2.0m apart. What is the field strength halfway between them?

3. The attempt at a solution

netEa= (9.0x10^9)(4.0x10^-6C) + (9.0x10^9)(8.0x10^-6C)
(2m)^2 (2m)^2

2. Feb 9, 2008

chrsr34

I assume by field strength, they want force, i.e. coulombs law.
Remember they want the strength at a point half way between them, half way being 1m from each of them. Use coulombs law for each charge with radius of 1m and then add the 2 forces together.

Chris

3. Feb 10, 2008

jalen

When you said with a radius of 1m you meant the (2m)^2 both become (1m)^2,right?

The answer in the text says 3.6x10^4N/C toward smaller charge but I got 1.08x10^5N/C if I used the(1m)^2 in my calculations......

netEa= (9.0x10^9)(4.0x10^-6C) + (9.0x10^9)(8.0x10^-6C)

(2m)^2 (2m)^2

4. Feb 10, 2008

chrsr34

Im not sure how you got that answer. 36000 is correct. Although I should correct myself by saying that you need to subtract the 2 forces because they are both positive, therefore oppose eachother.

So, ((9x10^9)(4x10^-6))/1m^2 = 36000
((9x10^9)(8x10^-6))/1m^2 = 72000

The force of q2 is canceling out the force of q1. So 72000-36000 = 36000 in the direction of q1.
Make sense?

Chris