What Is the Net Electric Force and Field on Charge 1?

[chef99]

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

F_net = √1.35N² + 1.35N²

F_net = 1.909N

F_net = 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​

 

[NoahCygnus]

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 $q_{1}$.

 

[ehild]

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.

​

 

[chef99]

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

 

[ehild]

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?

 

[chef99]

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?

 

[ehild]

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.

 

[chef99]

Yes, as the charge q1 is positive. If it was negative the direction of the electric field would be opposite to the force.

Thanks