Two particles having charges separated by a distance

  • Thread starter sonrie
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  • #1
sonrie
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Two particles having charges of 0.600 nC and 5.40 nC are separated by a distance of 1.30 m.

A.) At what point along the line connecting the two charges is the net electric field due to the two charges equal to zero?

the electric field is zero at a point _______ m from .600nC

i found out the answer to this to be .325m. by doing the following:

E due to q1 is equal in magnitude to E due to q2 but in opposite direction

As both q1 and q2 are positive, the field is zero at a point in between them

As distance from q1 is L , distance from q2 is (1.3 - L)

kq1/L^2 = kq2/(1.2 - L)^2

(1.3 - L) / L= sq rt (q2 / q1)

(1.3 - L) = L* sq rt (q2 / q1)

1.3 = L [1+sq rt (q2 / q1)]

L = 1.3 / [1+sq rt (q2 / q1)]

L = 1.3 / [1+sq rt (5.40 / 0.600)]

L = 1.3 / [1+sq rt ( 9.00 )]

L = 1.3 / [1+ 3.0]

L = 1.3 / [ 4.0]

L=0.325 meter from 0.550 nC

Where would the net electric field be zero if one of the charges were negative.
Enter answer as a distance from the charge initially equal 0.600 nC . This is where i get lost, i don't know what to do next. please Help!
__________________
 

Answers and Replies

  • #2
Phlogistonian
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You do it like you did the first one. Start with
[tex]\vec{E_1} + \vec{E_2} = 0[/tex]

Be careful about the vector directions.
 
  • #3
sonrie
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should the answer be the same but just negative since i using the same values as before?
 
  • #4
Phlogistonian
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When I wrote, "You do it like you did the first one," I meant, you start it the same way. The algebra is different.

I apologize; I should have been clearer.
 
  • #5
sonrie
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Thanks A Bunch!
 
  • #6
Phlogistonian
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You're welcome.
 

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