Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm currently studying electrodynamics where the following equation arose:

$0 = \frac{q}{\sqrt{R^2 + d^2 - 2Rd\cos\theta}} + \frac{q_p}{\sqrt{R^2 + d_p^2 - 2Rd_p\cos\theta}}$

where I need to solve for $q_p$ and $d_p$. The solution given is the following (factoring out $R^2$ out of the squareroot):

$0 = \frac{q/R}{\sqrt{1 + (d/R)^2 - 2 (d/R)\cos\theta}} + \frac{q_p/d_p}{\sqrt{1 + (R/d_p)^2 - 2(R/d_p)\cos\theta}} $ which lets us read out the solutions: $q/R = - q_p/d_p$ and $d/R = R/d_p$ which can be easily solved for $d_p$ and $q_p$.

My question is the following: Why can't I solve the equation starting from the first equation? The first equation is zero if we chose $q_p = -q$ and $d_p = d$. That solution doesn't make a lot of sense physically (it basically means putting to charges with opposing charges on top of each other), but why is it wrong mathematically? Or why do I lose solutions when I do it that way?

I hope someone can help.


edit: Physical context is the following: You have a charge on the $z$-Axis at distance $d$ to the origin. It has charge $q$. Around the origin is a conducting half-sphere of radius $R$. Since it is conducting, its potential on the surface has to be zero. To get the field in the space outside of the sphere we can use the method of image charges and place an image charge inside the sphere, such that the potential on the surface of the sphere is zero. This condition is the one written above. The superposition of two point charges has to be zero at $|\vec{x}| = R.$ $q_p$ is the charge of the image charge, and $d_p$ is its position on the z-Axis.

edit#2: The potential of a pointcharge at position $\vec{r}'$ is given by: $\phi(\vec{r}) = \frac{Q}{4 \pi \epsilon_0} \frac{1}{|\vec{r} - \vec{r}'|}$

share|cite|improve this question
Replace $d$ by $d_p$ in the numerator of the second fraction. – Tony Piccolo Aug 23 '13 at 9:41
You're right. Question still stands though. – user17574 Aug 23 '13 at 9:44
You want to solve a single equation with respect to two unknowns ... Can you give the physical context ? – Tony Piccolo Aug 23 '13 at 9:51
@TonyPiccolo I edited the inital post to clear the context up. – user17574 Aug 23 '13 at 9:57

I think the main point is that by setting $q_p = -q$ and $d = d_p$ you find only ONE solution, namely the trivial solution: it corresponds to a total zero charge placed at distance $d$, in this case the potential is zero not only on the sphere but everywhere. The method of images shows that there is also a non trivial solution. You could think of the problem backwards also: given two opposite charges in the $z$ axis, lying on the positive and negative semi-axis, what shape does the equipotential surface with potential $\phi =0$ have? You will find it is a sphere of radius $R$ only for the non trivial case.

share|cite|improve this answer

You can solve the first equation.

Square both sides of $$\frac{q}{\sqrt{R^2 + d^2 - 2Rd\cos\theta}}=-\frac{q_p}{\sqrt{R^2 + d_p^2 - 2Rd_p\cos\theta}}$$ and remove the denominators.
Considering the two sides of the equation so obtained as polynomials in $\cos \theta$, equate the coefficients to have the system $$\begin {cases} q^2(R^2+d_p^2)=q_p^2(R^2+d^2) \\ \\q^2d_p=q_p^2d \end {cases}$$ Dividing side by side the two equations, you have $$dd_p^2-(R^2+d^2)d_p+R^2d=0$$ which gives $$d_p=d \quad,\quad d_p=\frac {R^2}d$$If you substitute in the second equation of the system, you are done.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.