# Electric Field Due to Dipole Problem

## 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.)

Et= 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

2[kq/[d2/4 + x2]1/2]sinθ

This is not correct.

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Dick
Homework Helper

## 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.)

Et= 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

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.

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