Electric field from two point charges and then some

[rowkem]

Homework Statement

A point charge, q1, has a charge of 3.4x10^-9 C. A second point charge, q2, is place 2.1x10^-6 m to the right of q1; it has a charge of -10.2x10^-9C.

A) What is the magnitude of the field 8 mm to the left of the centre of the two charges?

B) You could also treat this collection of charges as a dipole added to a point charge. If you think of the charges this way, what would the charge on the point charge be?

C) What would the field be, due to just that point charge, 8 mm to the left of the charge?

Homework Equations

E=k((Q)/(r^2))

The Attempt at a Solution

A) I figured out part A to be -7.6x10^-3 N/C. I'm a little lost as to what this answer indicates about the direction of the electric field. Also, my answer is using 8x10^-3m for the r in both field calculations. Does it really matter if I consider the whole "from the centre of"?

B) Confused me completely. I'm not sure what it's asking me to do.

C) I'm assuming that once B is solved, I plug that value in for Q and will get another answer for the field?

 

[Delphi51]

I don't see how you could possibly get a magnitude without considering the directions!
Did you add or subtract the two fields? Better begin with a diagram. Take note of whether the 8 mm or the 2.1x10^-6 m is larger when you mark the point where we are calculating the field. Work out the field due to each charge separately and then decide whether you are adding or subtracting. The field direction is away from the positive charge and toward the negative one.

 

[nlightner]

Hello, thank you for your question. I would like to provide a response to your content.

Firstly, it is important to note that the electric field is a vector quantity, meaning it has both magnitude and direction. In this case, the negative sign in your answer for part A indicates that the direction of the electric field is towards the left. This is because the negative charge (q2) is closer to the point of interest than the positive charge (q1), resulting in a net electric field pointing towards the negative charge.

For part B, thinking of the charges as a dipole added to a point charge means that the two charges (q1 and q2) can be treated as one point charge located at the center of the dipole. The magnitude of this point charge can be calculated using the formula for electric dipole moment, p = qd, where q is the magnitude of one charge and d is the distance between the two charges. In this case, p = (3.4x10^-9 C)(2.1x10^-6 m) = 7.14x10^-15 Cm. This is the charge on the point charge that is equivalent to the dipole.

For part C, once you have the value for the point charge (from part B), you can use the formula for electric field (E = kq/r^2) to calculate the electric field at a distance of 8 mm to the left of the point charge. This will give you the electric field due to just the point charge, which can be compared to the answer you got for part A to see the effect of the dipole on the overall electric field.

In summary, it is important to consider both magnitude and direction when dealing with electric fields, and to understand the concept of equivalent charges in a system. I hope this helps and good luck with your homework!