Electric Field for Charge Distributions

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SUMMARY

The discussion focuses on calculating the electric field and net force for a system of three charges: q1 = +2.0 x 10-5 C, q2 = q3 = -3.0 x 10-5 C, with distances r31 = r21 = 2 m. The electric forces FE31 and FE21 are both calculated to be 1.35 N, leading to a net force (FNET) of 1.91 N at an angle of 45 degrees. The net electric field (εNET) acting on charge 1 is determined to be 9.55 x 104 N/C, also at 45 degrees. The calculations are confirmed to be correct, with suggestions for improving clarity and precision in significant figures.

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  • Learn about vector components and their role in calculating net forces and fields
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  • Review significant figures and precision in scientific calculations
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Jkalirai
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Homework Statement
a) Find the net force on charge 1.
b) What is the net electric field acting on charge 1?
Relevant Equations
F[SUB]E[/SUB] = k * q[SUB]1[/SUB] * q[SUB]2[/SUB] / r[SUP]2[/SUP]
ε = k * q / r^2
We are given: q1 = +2.0 x 10-5 C, q2 = q3 = -3.0 x 10-5 C, r31 = r21 = 2 m

a) We start by finding the electric force between q3 to q1 and q2 to q1

FE31 = k * q1 * q3 / r312

FE31 = (9.0 x 109 Nm2/C2) * (+2.0 x 10-5 C) * (3.0 x 10-5 C) / (2 m)2

FE31 = 1.35 N

FE21 = k * q1 * q2 / r212

Since q2 = q3 and r31 = r21

FE21 = 1.35 N

Since the force of gravity is assumed to be negligible, we know that FE21 = FNETx and FE31 = FNETy

We can calculate FNET from our components,

FNET = √ [(1.35)2 + (1.35)2]

FNET = 1.9092 N

The angle of our net force can be calculated as well,

tanΘ = (1.35 N / 1.35N)

Θ = tan-1 (1.35N / 1.35 N)

Θ = 45o

Therefore, the net force on charge 1 is equal to 1.91 N [N 45o E].

b) To find ε on charge 1 we solve for the components of ε.

We know that q2 = q3 and r2 = r3

Therefore, ε2 = ε3

ε2 = ε3 = k * q / r2

ε = (9.0 x 109 Nm2/C2) * (3.0 x 10-5 C) / (2 m)2

ε = 6.75 x 104 N/C

εNET = √ [(6.75 x 104)2 + (6.75 x 104)2]

εNET = 95459.4155 N/C

Our angle will be,

Θ = tan-1 (6.75 x 104 N/C / 6.75 x 104 N/C)

Θ = 45o

Therefore, our net electric field acting on charge 1 is 9.55 x 104 N/C [N 45o E].

I think my answer for part a) seems correct but for part b), I wanted to make sure it was right. It does seems to make sense when I think about it in my head, obviously the electric fields that are having an effect on charge A would be from charge B and C and their components would give me the net electric field. I've seen various different answers for part b) so I thought maybe I might be missing something. Anyways, I appreciate any input and thanks for giving this a read.
 

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You twice wrote ε = k * q / r, but correctly used ε = k * q / r2.
You could have made it a bit easier by doing b first then multiplying by q1.
You quote too many significant figures in the answers, given the precisions of the original data.
Other than that, looks fine.
 
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