Electric Field Due to Dipole Problem

[Renaldo]

Homework Statement

Consider the following figure.

For the electric dipole shown in the figure, express the magnitude of the resulting electric field as a function of the perpendicular distance x from the center of the dipole axis. (Use the following as necessary: k, q for the charges, x, and d.)

Homework Equations

E_t= E₁ + E₂

E = k|q|/r²

The Attempt at a Solution

The x component of the electric fields will cancel out, leaving only the y-components. Adding the two vector fields in the y-direction:

[k|q|/r²]sinθ + [k|q|/r²]sinθ = 2[k|q|/r²]sinθ

r^2 = [d/2 + x^2]^1/2

My final answer:

2[kq/[d²/4 + x²]^1/2]sinθ

This is not correct.

 

[Dick]

Homework Statement

Consider the following figure.

For the electric dipole shown in the figure, express the magnitude of the resulting electric field as a function of the perpendicular distance x from the center of the dipole axis. (Use the following as necessary: k, q for the charges, x, and d.)

Homework Equations

E_t= E₁ + E₂

E = k|q|/r²

The Attempt at a Solution

The x component of the electric fields will cancel out, leaving only the y-components. Adding the two vector fields in the y-direction:

[k|q|/r²]sinθ + [k|q|/r²]sinθ = 2[k|q|/r²]sinθ

r^2 = [d/2 + x^2]^1/2

My final answer:

2[kq/[d²/4 + x²]^1/2]sinθ

This is not correct.

For one thing r^2=x^2+(d/2)^2. You've got an extra square root. For another, you should be able to express sinθ in terms of x and d as well.

 

[Renaldo]

For one thing r^2=x^2+(d/2)^2. You've got an extra square root. For another, you should be able to express sinθ in terms of x and d as well.

Yes, I noticed that extra square root. When I fix things up, and set sinθ equal to d/2r:

It's a lot of algebra, but it comes out to kqd/[(d²/4)+x²)]^3/2. Does that look right?

 

[Dick]

Yes, I noticed that extra square root. When I fix things up, and set sinθ equal to d/2r:

It's a lot of algebra, but it comes out to kqd/[(d²/4)+x²)]^3/2. Does that look right?

Looks ok to me.

 

[Renaldo]

Looks ok to me.

All right. Thanks. Webassign doesn't like the answer but I feel confident in the math. Could be a problem with Webassign.