Net electric force on a charge

  • Thread starter chef99
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Homework Statement


A) Determine the net force on charge 1.
B) What is the net electric field acting on charge 1?



Homework Equations


Fnet
Enet

The Attempt at a Solution



Let North and East be positive.

Because of symmetry, F21 = F31

F2 = F3

F2 = (9.0x109)(2.0x10-5(3.0x10-5) / (2.0m)2

F2 = 1.35N

Fnet = √1.35N2 + 1.35N2

Fnet = 1.909N

Fnet = 2N

For the angle of the electric force:

tan-1 = 2/2

= 45º

therefore the net electric force on charge 1 is 2N [N 45º E].


B) Enet

Let right and up be positive.

Because of symmetry,

r = √22 + 22

= 2√2

Enet = 2E1cos45º

Enet = 2(kq1 / r2 cos45º

Enet = 2(9.0x109)(3.0x10-5) / (2√2)2 cos45º

Enet = 47729.7 N/C

Enet = 4.7 x104 N/C




Any input is appreciated as I still am not completely confident that I am doing these questions right. Thanks​
 

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Answers and Replies

  • #2
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The force calculation part is correct but I don't get what you have done with the electric field. I think there is a problem with your notation, electric field ##\vec{E_{1}}## due to ##q_{1}## at the location where ##q_{1}## is located will be infinite, you don't have to consider charge ##q_{0}## in the evaluation of electric field at that location. The electric fields of ##q_{2}## and ##q_{3}## will add according to superposition principle. Use the definition of electric field , ##\vec{E} = \Sigma\vec{F}/q_{0}## , to calculate electric field at the location of ##q_{1}##.
 
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  • #3
ehild
Homework Helper
15,543
1,909
B) Enet

Let right and up be positive.

Because of symmetry,

r = √22 + 22

= 2√2

Enet = 2E1cos45º
Recall the definition of the electric field strength: it is the electric force acting on unit positive charge.
You need the electric field at the position of charge 1, which is simply the electric force acting on the charge 1 divided by q1. E=F/q1.



 
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  • #4
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Recall the definition of the electric field strength: it is the electric force acting on unit positive charge.
You need the electric field at the position of charge 1, which is simply the electric force acting on the charge 1 divided by q1. E=F/q1.



So

Enet= F/q1

= 2N/C/ 2.0x105

=100000

The net electric field acting on charge 1 is

1.0x105N/C
 
  • #5
ehild
Homework Helper
15,543
1,909
So

Enet= F/q1

= 2N/C/ 2.0x105

=100000

The net electric field acting on charge 1 is

1.0x105N/C
OK, but what is the direction of the net electric field?
 
  • #6
75
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OK, but what is the direction of the net electric field?
would I state: The net electric field acting on charge 1 is 1.0 x105N/C [N 45º E] --as I have already determined the direction of the electric force or would I have to calculate it differently for the electric field? i.e. is the direction of the electric field the same as the electric force?
 
  • #7
ehild
Homework Helper
15,543
1,909
would I state: The net electric field acting on charge 1 is 1.0 x105N/C [N 45º E] --as I have already determined the direction of the electric force or would I have to calculate it differently for the electric field? i.e. is the direction of the electric field the same as the electric force?
Yes, as the charge q1 is positive. If it was negative the direction of the electric field would be opposite to the force.
 
  • #8
75
4
Yes, as the charge q1 is positive. If it was negative the direction of the electric field would be opposite to the force.
Thanks
 

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