Is this the correct equation to find the electric field at point p from a dipole?

[novelriver]

Is this the correct equation to find the electric field at point p from a dipole?

 

[BvU]

Hello Novel,

You don't want to delete the template; it's very useful for you as well as for us. See the guidelines.

Answer to your question: No. $\ \vec E\$ is a vector. What does your E describe, you think ?

 

[novelriver]

Hello Novel,

You don't want to delete the template; it's very useful for you as well as for us. See the guidelines.

Answer to your question: No. $\ \vec E\$ is a vector. What does your E describe, you think ?

Thanks for your reply. I'll make sure to not delete the template next time. I think my E describes the magnitude of the electric field. After calculating the magnitude, I can decide if it's positive or negative. So in this case, E is negative, because E is going down, away from positive and toward negative. Is this right?

 

[BvU]

I'll give you some leeway because you are new here (let's hope I don't get chastized for that).
Also because I think you have a fair idea what you are doing, but you stumble because you are going too fast.

Again, $\vec E$ is a vector. I've drawn the two contributions from the +Q and the -Q in the figure.
There are no other contributions, so the field at P is the sum of these two. The vector sum, that is. Your job to do this vector addition. Andf yes, x/r appears in there (not x/R but x/r; I don't see or know of R in your post. Work accurately ). And yes, it's downwards. Easy exercise, but a good vehicle to learn to work systematically.

 

[novelriver]

I'll give you some leeway because you are new here (let's hope I don't get chastized for that).
Also because I think you have a fair idea what you are doing, but you stumble because you are going too fast.

Again, $\vec E$ is a vector. I've drawn the two contributions from the +Q and the -Q in the figure.
There are no other contributions, so the field at P is the sum of these two. The vector sum, that is. Your job to do this vector addition. Andf yes, x/r appears in there (not x/R but x/r; I don't see or know of R in your post. Work accurately ). And yes, it's downwards. Easy exercise, but a good vehicle to learn to work systematically.View attachment 105666

I think I understand.

E = kQ/r² * cos(theta) because the y-components cancel out and we just want to get the x-component. I'll call the horizontal distance d (in a real problem it would be given or I could find it with trig), so cos(theta) = d/r. Therefore E = kQ/r² * d/r = kQd/r³.

 

[novelriver]

I think I understand.

E = kQ/r² * cos(theta) because the y-components cancel out and we just want to get the x-component. I'll call the horizontal distance d (in a real problem it would be given or I could find it with trig), so cos(theta) = d/r. Therefore E = kQ/r² * d/r = kQd/r³.

I just realized my mistake. It's the x-components that cancel, not the y. So E = kQ/r² * sin(theta) = kQ/r² * x/r = kQx/r³ and then multiply by 2 because there are two charges acting on P in the same direction. So E = 2kQx/r³.

 

[BvU]

Looks good to me. x and y are a bit confusing here because of the x's in the figure.

 

[novelriver]

Looks good to me. x and y are a bit confusing here because of the x's in the figure.

Right, I realized that I should have chose a different variable name than x. Thank you for your help! I appreciate it!