I have two spheres of metal, each of radius $1m$, these spheres are effectively an infinite distance apart.

I now connect a $1V$ power supply between these two spheres: (and wait an infinite amount of time, for the charges to reach equilibrium.)

   O----------------| 1V |----------------O
Sphere 1         Power Supply          Sphere 2
  1. What property of the metal from which the spheres are made determines the magnitude of the charge on the spheres?
  2. How can the magnitude of the charge on one of the spheres be calculated?
  3. If one of the spheres had its mass doubled what would be the charge on the spheres be then?

(I am hoping to discover a new equation which will allow me to solve a more in depth problem, which I dreamt up whilst sitting on the bus)


I have now discovered what I think is the equation which I was looking for: $Q=4\pi \epsilon_0 R V$ for a single isolated sphere. I therefore (by intuition) think that the equivalent for the 2 sphere set-up is $Q=2\pi \epsilon_0 R V$

Therefore the charge I was looking for would be $Q=2\pi \epsilon_0 \times 1V \times 1m=5.56325...\times10^{-10}C$

  • 1
    $\begingroup$ The magnitude of the charge on either sphere just depends on the voltage and the capacitance between the spheres. And the capacitance just depends on the geometry of the setup (i.e., radii of the spheres and the distance between them). Doesn't matter what metal(s) they are made out of. $\endgroup$ – user93237 Mar 10 '16 at 20:48
  • $\begingroup$ @SamuelWeir So what if the spheres are an "infinite" distance apart? $\endgroup$ – o.comp Mar 10 '16 at 21:14
  • $\begingroup$ As the distance between two spheres of fixed size goes to infinity, the capacitance between them goes to zero. $\endgroup$ – user93237 Mar 10 '16 at 22:49
  • $\begingroup$ @SamuelWeir Surely at some point the internal repulsion of the electrons in the negative sphere will reach equilibrium with the powers supply, even if this is no longer technically capacitance. $\endgroup$ – o.comp Mar 11 '16 at 6:53
  • $\begingroup$ $Q=4\pi \epsilon_0 R V$ is correct and $2π \epsilon RV$ is wrong $\endgroup$ – Anubhav Goel Mar 16 '16 at 4:51

Let the distance between the two spheres be $d$ and let one of them have a charge of $Q$ and the other, $-Q$. The net field at a point at a distance $r$ from the centre of the positively charged sphere is the sum of fields due to both charges, and the direction of field is same for both direction, i.e. towards the negatively charged sphere. Thus the field is $\displaystyle E(r) = \dfrac {Q}{4\pi\varepsilon_0}\left(\dfrac 1{r^2} +\dfrac1{(d-r)^2}\right)$, and the potential difference between the two spheres is: $$ \begin{align}V &= \int_R^{d-R}\dfrac {Q}{4\pi\varepsilon_0}\left(\dfrac 1{r^2} +\dfrac1{(d-r)^2}\right) \\[2ex] &=\dfrac {Q}{4\pi\varepsilon_0R}\left(\dfrac {-1}{r} +\dfrac1{d-r}\right)_{R}^{d-R}\\[2ex] &=\dfrac {2Q}{4\pi\varepsilon_0R}\left(\dfrac{d-2R}{d-R}\right) \end{align}$$ $$\therefore Q = 2\pi\varepsilon_0RV\cdot\dfrac{d-R}{d-2R}$$ Taking the limit $d\to\infty$, we get: $$Q=2\pi\varepsilon_0RV$$ Please correct me if i am wrong. Hope this helps. Cheers!

Note: A mention of thanks to @Anubhav Goel, for helping me realize my previous approach using direct expressions for potential of charged sphere wasn't that good, and for pointing out the mistake in my integral. So i have edited the answer to show another method using electric field, and corrected the integral.

| cite | improve this answer | |
  • $\begingroup$ How did you take out PD b/w two spheres? $\endgroup$ – Anubhav Goel Mar 16 '16 at 4:47
  • $\begingroup$ I used the formula for the electric potential of a solid sphere, and took the scalar sum of the potential due to each sphere at each point. $\endgroup$ – FreezingFire Mar 16 '16 at 5:38
  • $\begingroup$ @AnubhavGoel, i suppose that method may not be the best way here. So i have edited my answer to solve the question using electric field. Thank you for your interest! :) $\endgroup$ – FreezingFire Mar 16 '16 at 5:58
  • $\begingroup$ $R$ is the radius of the sphere, that is $\text {1m}$. Also, ∞/∞ form does not mean limit does not exist. Also my limit evaluates correctly. $\endgroup$ – FreezingFire Mar 16 '16 at 10:02
  • $\begingroup$ @AnubhavGoel And many apologies, my integration was wrong. I have corrected it now! Thanks again! $\endgroup$ – FreezingFire Mar 16 '16 at 10:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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