Electric Field for Charge Distributions

In summary, we calculated the electric force between q3 and q1 to be 1.35 N and the electric force between q2 and q1 to also be 1.35 N. The net force on charge 1 is 1.91 N at an angle of 45 degrees. To find the net electric field on charge 1, we found the components of the electric field to be 6.75 x 104 N/C and the net electric field to be 9.55 x 104 N/C at an angle of 45 degrees. The calculations for the net electric field were done by first finding the components of the electric field and then multiplying by the charge of q1. The precision of the original data
  • #1
Jkalirai
4
1
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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  • #2
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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FAQ: Electric Field for Charge Distributions

What is an electric field?

An electric field is a physical quantity that describes the influence that an electric charge has on other charges in its vicinity. It is represented by a vector that indicates the direction and strength of the force that would be exerted on a positive test charge placed in the field.

How is an electric field calculated for a point charge?

The electric field at a point in space caused by a point charge can be calculated using the equation E = kQ/r^2, where k is the Coulomb's constant, Q is the magnitude of the point charge, and r is the distance between the point charge and the point at which the field is being calculated.

How do you calculate the electric field for a continuous charge distribution?

To calculate the electric field for a continuous charge distribution, the distribution must be divided into small elements and the electric field at each element must be calculated using the equation E = kΔq/r^2. The electric fields from each element can then be added together to find the total electric field at a given point.

What is the relationship between electric field and electric potential?

The electric field and electric potential are closely related. Electric potential is defined as the electric potential energy per unit charge, and it is directly proportional to the electric field. In other words, the electric field is the gradient of the electric potential.

How does the shape of a charge distribution affect the electric field?

The shape of a charge distribution can greatly affect the electric field it produces. For example, a spherical charge distribution has a uniform electric field in all directions, while a charged rod will have a stronger field at its ends compared to its center. The direction and strength of the electric field can also be affected by the orientation of the charge distribution.

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