The potential of a hemisphere at the centre with constant surface charge density $\sigma $ is given by $\frac { \sigma R }{ 2\epsilon } $ where $R$ is the radius of the hemisphere. The magnitude of the electric field for the same configuration is given by $\frac { -\sigma }{ 4\epsilon } $.

I've got the potential by doing a surface integral and the field by integrating rings that make the hemisphere.

But why can't we use $E=-\nabla V$ arriving at $\frac { -\sigma }{ 2\epsilon } $ for the field.


  • 1
    $\begingroup$ If the potential $\;V(\mathrm A)\;$ at a point $\;\mathrm A\;$ is $\;V(\mathrm A)=5$ Volts could we deduce from this what is the field $\;E(\mathrm A)\;$ at this point or we must know the rate of change of $\;V\;$ along 3 not co-planar directions (for example $\;\mathrm dV/\mathrm dx$,$\;\mathrm dV/\mathrm dy$,$\;\mathrm dV/\mathrm dz$) in the vicinity of $\;\mathrm A$??? If I tell you that I stand on a hillside at a height $\;h=135m\;$ from the sea DO I give you any information about the slope in any direction of the hill at this point ??? $\endgroup$ – Frobenius Aug 17 '18 at 11:49

Given you have evaluated the potential as number, you cannot take the derivative of this number w/r to anything.

To proceed with the gradient, you need to find $V$ as a function of the position (by symmetry you can restrict to the position on the axis), i.e. $V(z)$ and then $\vec E=-\hat z \frac{\partial V}{\partial z}$ since by symmetry the field cannot have components along any other axis.


In your specific example, you can argue by symmetry that the field $\vec E$ for a point on the axis of symmetry will be along $\boldsymbol{\hat z}$. Hence, you compute the potential for any point $V(z)$ on the symmetry axis (rather than just at the center of the hemisphere), you can compute $E_z=-\frac{\partial V}{\partial z}$.

With reference to the geometry below (where the radius of the sphere is $a$ rather than your $R$)

enter image description here

the expression for the potential at a point $P$ located at $(0,0,z)$ on the symmetry axis with $z>0$ is $$ V(z) = \frac{Q}{4\pi\epsilon_0}\frac{(a+z -\sqrt{a^2+z^2})}{az}\, . \tag{1} $$ so you can easily recover the field from that.

Obtaining (1) is reasonably straightforward although there is a possibly tricky integral which you might have to look up.

It might also be useful to know that, for $z$ small but positive, a series expansion of $\frac{(a+z -\sqrt{a^2+z^2})}{az}$ shows that the potential remains finite at $z=0$ and in fact has a part linear in $z$, so the field will also be finite at $z=0$.

[Please note that I copied (1) from an old textbook: I do not guarantee it is correct. You will have to check this yourself but given the behaviour for small $z$ Eq.(1) seems like a reasonable expression.]

  • $\begingroup$ The answer depends on the radius of the sphere in general and in this case is R. So does it make a difference? $\endgroup$ – AlphaBaal Oct 16 '17 at 16:25
  • 1
    $\begingroup$ @Alpha7200 Not really. Basically $\partial V/\partial z \approx (V(z+\Delta z)-V(z))/\Delta z$ so you really need $V$ at $z$ and $V$ at $z+\Delta z$. Even if you fix your $z$ to be at the center of the hemisphere, you still need $V$ at position $\Delta z$ away from the center of the hemisphere. What you need to do is to compare $V$ for fixed $R$ at two different points: $z$ and $z+\Delta z$. $\endgroup$ – ZeroTheHero Oct 16 '17 at 17:11
  • $\begingroup$ So how do I go about now finding the field from the potential? $\endgroup$ – AlphaBaal Oct 16 '17 at 18:17
  • $\begingroup$ @Alpha7200 added some material for guidance. $\endgroup$ – ZeroTheHero Oct 16 '17 at 19:35

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.