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Electric Field's Direction

  • #1

Homework Statement



A negative point charge q_1 = -4.00 nC is on the x-axis at x = 0.60 m. A second point charge q_2 is on the x-axis at x = -1.20 m.

a. What must the sign and magnitude of q_2 be for the net electric field at the origin to be 50.0 N/C in the +x-direction?

b. What must the sign and magnitude of q_2 be for the net electric field at the origin to be 50.0 N/C in the -x-direction?



Homework Equations



E = k*q/r^2

The Attempt at a Solution




a. The charge must be negative for subtraction to occur.

(-q2) O__<____. _>__O (-q1)
-----E2 = (-)------E1= (+)
negative diagram
. = field point (0,0)
O = particle
Arrow indicates field direction based on positive test charge

E1 = (8.988*10^9)*(-4.00*10^-9 C)/(.60 m)^2 = 99.8667 N/C (+)

99.8667 N/C + E2 = 50 N/C
E2 = -49.8667 C

q_2 = r^2*(49.86667)/(8.988*10^9) = -7.9893*10^-9 C ?????




b. E1 = (8.988*10^9)*(-4.00*10^-9 C)/(.60 m)^2 = 99.8667 N/C (+)

E2 = -50 N/C – 99.8667 = -149.8667 C

q_2 = [(r^2)149.8667 N/C]/k = [(r^2)149.8667 N/C]/k = (1.2^2)*149.8667/(8.988*10^9) = -2.40*10^-8 C ??????

Thanks.
 

Answers and Replies

  • #2
berkeman
Mentor
56,476
6,380
For a), why do you say "The charge must be negative for subtraction to occur"? E field points from + charge to - charge, and a) wants a net positive E field at the origin, doesn't it? Maybe I'm misreading something.
 
  • #3
72
0
u should know that q1 and q2 exert on the origin ... so there is 2 forces what means there is 2 electric fields..
the net electric fields should be =50 and in the direction of the force exert by q2 on the origin...u have this condition and the equations ...u should solve it i think
tr to get the net force on the origin assuming theer is test point charge. then using the condition u have and the values u will have the answer.
 
  • #4
BobG
Science Advisor
Homework Helper
185
80
Part a is correct.

You can look at this by proportions as a sanity check. q2 is twice as far away as q1, so if it had an equal charge, the strength of its field at the origin would be 1/4 as strong as q1's. Since the field was half as strong as it would be with q1 alone, the charge on q2 has to be about twice as strong as q1.
 
  • #5
72
0
hey BobG is what i said true? just to check if i know the concept
 
  • #6
BobG
Science Advisor
Homework Helper
185
80
hey BobG is what i said true? just to check if i know the concept
That's a tough question.

You could insert a test point of some mass (1kg to keep things simple), but you don't really need it. You'd be multiplying everything by your test mass, including your field at the origin.
 
  • #7
Is my answer for Part B incorrect?
 
  • #8
BobG
Science Advisor
Homework Helper
185
80
No, I think it's correct. I just didn't look at it. A quick check comparing the proportions looks right.
 

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