Electric field points with 3 charges

  • Thread starter rinarez7
  • Start date
1. The problem statement, all variables and given/known data
Consider 3 charges along a horizontal line as shown:
8.72e-6 C (+) ___4.16cm____2.45e-6 C (+)______3.18cm_____-1.46e-6C (-)
The acceleration due to gravity is 9.8 m/s^2
The Coulomb Constant is 8.98755e 9 Nm^2/ C^2
What is the electric field at a point 2.18 cm to the left of the middle charge? Answer in units fo N/C

2. Relevant equations
Q1= 8.72e-6 C , r= .0198m (.0416m-.0218m)
Q2= 2.45e-6 C, r= .0218m

E1= k Q1 / r^2
E2= k Q2/ r^2

3. The attempt at a solution
I converted all of my units first to C and meters.
Then I solved for E1 and E2 separately.
Once I had E1 and E2, I added the two together. But I can't seem to get the right answer.
Am I supposed to subtract them instead becasue Q1 and Q2 are both positive and therefore will repel each other?


Homework Helper
don't you have 3 charges?

draw a picture & look at the relative directions of the field form each charge to decide whether to add or subtract
Last edited:
so I calculated E1 = 1.999e8, E2 as 4.633e5, and E3 as -3.657e6
Am I correct to say then that, E1 and E3 are negative (pointing to the left) and E2 is positive?
E1(+)<----X----->E2 (+)-----><------E3 (-)


Homework Helper
sounds good, though unless its already given, you can define positive or negative anyway you like - you should specify this direction with your answer
Thank you!!!

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