Calculating Net Force and Electric Field Using Coulomb's Law - Homework Question

[kurt1992]

Homework Statement

This is the charge distribution:B
|
|
| 2.0 m
|
|
A------------C
2.0 m

A = 2.0*10^-5 C
B = -3.0*10^-5 C
C = -3.0*10^-5 C

a) Find the net force on the charge

b) what is the net electric field acting on the charge

Homework Equations

FE=kq1q2/r^2

electric field = Fnet/q

The Attempt at a Solution

a) The charges acting on A are equal and at equal distance so they are then same.

F_E=(9.0*10^9)(2.0*10^-5)(-3.0*10^-5)/2^2

F_E=5.4/4

F_E=1.35 N

Each charge has a net force of 1.35 Newtons on charge 1. However, force is a vector quantity so the vectors have to have the same direction in order to add them. (The forces are the same so we can simply multiply the hypotenuse of the electric force by 2 to find the net force on particle 1.

(F_E/cos45)(2)=Fnet

3.81838=Fnet

Direction N45E

(is this correct? my books solution to the problem involves adding x and y components which make a very confusion solution.)

b)

ε=3.81838/(2.0*10^-5)
ε=190919 N/C

 

[haruspex]

(F_E/cos45)(2)=Fnet

That's not right. Is that what you meant to write?

 

[kurt1992]

yes, i thought adding the vectors together at 45 degrees would be the same as adding x and y components and then using trigonometry to find the magnitude.
Is it not equivalent?

 

[haruspex]

yes, i thought adding the vectors together at 45 degrees would be the same as adding x and y components and then using trigonometry to find the magnitude.
Is it not equivalent?

Yes, if done correctly. Your answer was almost right, so I wondered if it was just a typo. If it's what you meant to post, please post your full working.

 

[Nickie]

The solution provided by the book is correct, but your approach is also correct. In this case, since the charges are equal and at equal distances, the net force on charge A will be in the direction of the line connecting the charges B and C, which is N45E. Therefore, your calculation of the net force on charge A is correct.

For part b), the net electric field acting on charge A will also be in the direction of N45E, but you have calculated the magnitude of the electric field instead of the net electric field. The net electric field is the sum of the individual electric fields from charges B and C, so you need to add the x and y components of the electric fields using vector addition.

Ex: The x component of the electric field from charge B is given by:

E_Bx = kq/r^2 * cos(theta)

where theta is the angle between the line connecting charges B and A and the x-axis. In this case, theta is 45 degrees, so cos(theta) = cos(45) = 1/sqrt(2).

Similarly, the y component of the electric field from charge B is given by:

E_By = kq/r^2 * sin(theta)

where sin(theta) = sin(45) = 1/sqrt(2).

Using these equations, you can calculate the x and y components of the electric field from charge B. Repeat this for charge C and then add the x and y components of the electric fields to find the net electric field acting on charge A.

I hope this helps! Keep up the good work with Coulomb's Law!