Calculation of the field due to a dipole at an arbitrary point

• Hamiltonian
In summary, the conversation discusses deriving the electric field at a point P using the equations E=−∇V and E=(kP/r3)(1+3cos3θ)1/2 in polar coordinates. The individual components of the electric field due to positive and negative charges at point P are calculated and then added to find the net electric field. To eliminate the variables alpha and beta, the law of sines and cosines are applied to the triangles formed by the charges. The final expression for the net electric field is a scalar, as the question only asks for the magnitude of the electric field.
Hamiltonian
Homework Statement
Calculate the magnitude of the field due to a dipole at a point P which is at a distance ##r## from the midpoint of the two charges and makes an angle ##\theta## with the dipole moment Vector ##P##. the distances between the two opposite charges is ##a## and ##(a<<r)##.
Relevant Equations
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I know how to derive field using ##E = -\nabla V## in polar coordinates and doing so gave me $$E = (kP/r^3)(1 + 3cos^3\theta)^{1/2}$$

now I am trying to derive ##E## at point P using the fields produced by +ve and -ve charge respectively and taking components of each along the radial direction along ##r## and perpendicular to ##r##. I assumed angles ##\alpha## and ##\beta## with the perpendicular to the radial direction in hopes they get eliminated when finding ##E_{net}##.

$$E_+ = \frac {kq}{r^2} (1+ (2a/r) cos\theta)$$
$$E_- = \frac {kq}{r^2} (1- (2a/r) cos\theta)$$
here ##E_+## and ##E_-## are fields due to the -ve and +ve charges at point P.
$$E_r = E_+ sin\beta - E_- simn\alpha$$
$$E_{r'} = E_+ cos\beta + E_- cos\alpha$$
here ##E_r## is the componetnt of ##E_{net}## at P along ##r## and ##E_{r'}## is the component of ##E_{net}## perpendicular to ##r##.
$$| E_{net}| = \sqrt{(E_r)^2 + (E_{r'})^2}$$
I am not able to eliminate ##\alpha## and ##\beta## from the final expression for ##E_{net}##.

Last edited:
Delta2
Hamiltonian299792458 said:
I know how to derive field using E=−∇V in polar coordinates and doing so gave me E=(kP/r3)(1+3cos3θ)1/2
This doesn’t seem right. You have given a scalar, but the electric field is a vector field. Please show your work

Orodruin said:
This doesn’t seem right. You have given a scalar, but the electric field is a vector field. Please show your work
I forgot to put an arrow on top of ##E##
$$\vec E = -\nabla V$$

although my doubt wasn't related to this, this is what I did
$$V_p = (kPcos\theta)/r^2$$
$$|\vec E| = \sqrt{(-\frac{\partial V}{\partial r})^2 + (-\frac{1}{r}\frac {\partial V}{\partial \theta})^2} = \frac{kP}{r^3}\sqrt{1+3cos^2\theta}$$

.
Hamiltonian299792458 said:
View attachment 281253
$$E_+ = \frac {kq}{r^2} (1+ (2a/r) cos\theta)$$
$$E_- = \frac {kq}{r^2} (1- (2a/r) cos\theta)$$
I don't think the factors of 2 are correct in the right-hand sides of these two equations. Post your work if you would like us to check it.

$$E_r = E_+ sin\beta - E_- sin\alpha$$
$$E_{r'} = E_+ cos\beta + E_- cos\alpha$$

I am not able to eliminate ##\alpha## and ##\beta##
Try to relate ##\beta## and ##\theta##. Hint: apply the law of sines to one of the triangles in your figure.

Hamiltonian
TSny said:
.
I don't think the factors of 2 are correct in the right-hand sides of these two equations. Post your work if you would like us to check it.
the distance between the two opposite charges is ##2a##
applying the cosine rule to both the triangles gives
$$r_+^2 = r^2 + a^2 -2arcos\theta \approx r^2 -2arcos\theta$$
$$r_-^2 = r^2 +a^2 -2arcos\theta \approx r^2 +2arcos\theta$$

also after applying the sine rule to both the triangles and making a few approximations I was able to get the required answer thanks!

the distance between the two opposite charges is ##2a##
OK. In the original post, the distance is given to be ##a## rather than ##2a##. That accounts for why I thought you were off by a factor of 2. I'm glad everything worked out.

Hamiltonian
Hamiltonian299792458 said:
I forgot to put an arrow on top of ##E##
$$\vec E = -\nabla V$$

although my doubt wasn't related to this, this is what I did
$$V_p = (kPcos\theta)/r^2$$
$$|\vec E| = \sqrt{(-\frac{\partial V}{\partial r})^2 + (-\frac{1}{r}\frac {\partial V}{\partial \theta})^2} = \frac{kP}{r^3}\sqrt{1+3cos^2\theta}$$
My concern was not the missing arrow, it was the fact that your computed ##E## is a scalar field and not a vector.
Hamiltonian299792458 said:
and doing so gave me $$E = (kP/r^3)(1 + 3cos^3\theta)^{1/2}$$

Orodruin said:
My concern was not the missing arrow, it was the fact that your computed ##E## is a scalar field and not a vector.
the question states calculate the magnitude of the electric field due to an electric dipole at a point ##P##

$$|\vec E| = \frac{kP}{r^3}(1+ 3 cos^2\theta)^{1/2}$$

1. What is a dipole?

A dipole is a pair of equal and opposite charges separated by a small distance. It is a fundamental concept in electromagnetism and is used to describe the behavior of electric fields.

2. How is the field due to a dipole calculated?

The field due to a dipole at an arbitrary point is calculated using the formula E = (k*q*d)/r^3, where E is the electric field, k is the Coulomb's constant, q is the magnitude of the charges, d is the distance between the charges, and r is the distance from the dipole to the point of interest.

3. What is an arbitrary point?

An arbitrary point is any point in space that is not specifically defined or located. In the context of calculating the field due to a dipole, it refers to a point that is not on the dipole itself, but rather at some distance away from it.

4. Can the field due to a dipole be negative?

Yes, the field due to a dipole can be negative. This occurs when the point of interest is located on the opposite side of the dipole from the positive charge. In this case, the electric field will point in the opposite direction of the dipole moment.

5. How does the distance from the dipole affect the strength of the field?

The strength of the field due to a dipole decreases as the distance from the dipole increases. This is because the field follows an inverse square law, meaning that the field strength is inversely proportional to the square of the distance from the source.

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