# Charged Hemisphere

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.

Thanks.

• 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 ??? – 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.

Edit:

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 $R$)

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.]

• The answer depends on the radius of the sphere in general and in this case is R. So does it make a difference? – AlphaBaal Oct 16 '17 at 16:25
• @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 $z+\Delta z$. – ZeroTheHero Oct 16 '17 at 17:11
• So how do I go about now finding the field from the potential? – AlphaBaal Oct 16 '17 at 18:17
• @Alpha7200 added some material for guidance. – ZeroTheHero Oct 16 '17 at 19:35