Dipole potential and sphere grounded

Find the electrostatic potential arising from an electric dipole of magnitude $$d$$ situated a distance $$L$$ from the center of a grounded conducting sphere of radius $$a$$. Assume the dipole axis cross the sphere's center.

Honestly, this is not a hard question, it is more about be patient, solve some cosine laws and know the right approximations that is needed to use.

But the problem is not how to solve it, i have actually tried by another way than calculate the potential of the dipoles separated and sum it (including the potential of the net charge that will arise).

I have considered to use the formula $$\vec p = \int \rho \vec r d^3x$$ (in this case, basically $$\sum q z$$).

Assuming, first, that the distance between the dipole's charges is $$2l$$.

So, considering the sphere's center as the origin, the dipole we will get is (without any approximation) $$\vec p = (2ql - \frac{qR^3}{(L+l)^2} + \frac{qR^3}{(L-l)^2} )\space \hat z$$

the net charge will be

$$\delta q = \frac{2qR}{l^2-L^2}$$

Here comes the problem:

Isn't the potential of a system, until the dipole order, $$\phi = \frac{k \space q_{net} }{r} + \frac{k \space \vec p \vec r}{r^3}$$

$$?$$

Using this as the potential, we get a answer that involve just $$cos(\theta)$$. But the right answer is different, and i understand how the answer the book got was arrived. My question is why i can't use the one i posted here?

• I think you can use your answer. No problem to me. Which book do you use ? Jun 19 at 14:28

I noticed a little mistake (which is even detectable by dimensional analysis):

$$q_{net}= \frac{2qRl}{L^2-l^2}$$

Coming back to your question, how are the two answers different? The exact answer is the field created by the $$4$$ point charges (including the $$2$$ induced,virtual ones) so it has multipolar contributions at every order. However, each term should agree whether you expand out the exact solution in multipole term or whether you calculate the multipole moment of the distribution and plug in the formula of the resulting multipole field.

By the way, since the net charge is non trivial, you can kill the dipolar term by choosing the origin of the multipole expansion as the center of charge, ie at the position $$\vec p/q_{net}$$. Hope this helps and tell me if you find some mistakes.

• Not sure what mistake are you talking about in first line. (we have [C][m^2]/[m^2] = [C]). "How are the answer different?" That is what i am trying to get too, but i think i have figured out. Maybe the answer the book given is, actually, the exact answer, and mine answer is just approximation? I will add the answer the book gives at my post above
– LSS
Apr 27, 2022 at 11:59
• I was talking about your $\delta q$ formula which I understood as the total charge. It does not have the unit of charge and it also has the wrong sign. As I mentioned, the solution of the book is exact. In order to retrieve the two terms you wrote out, you need to do the multipole expansion. Mathematically, you need to expand the formula as a power series of $\cos \theta$ and the first two terms correspond to the ones you wrote above.
– LPZ
Apr 27, 2022 at 13:43