Calculating Electric Field Strength between Charges

AI Thread Summary
Two charges, +4.0x10^-6 C and +8.0x10^-6 C, are positioned 2.0m apart, and the electric field strength halfway between them needs to be calculated. The correct method involves using Coulomb's law to determine the forces exerted by each charge at a distance of 1m. The calculations yield a net electric field strength of 36,000 N/C directed towards the smaller charge. It is important to note that the forces from both charges oppose each other, necessitating subtraction to find the net effect. The final answer confirms the correct approach to calculating electric field strength in this scenario.
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Homework Statement



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

Homework 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?

The Attempt at a Solution



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

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
 
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