Help me calculate the field strength at the surface of the sphere?

hi guys any help me with this?

A metal Spher, suitably insulated from earth, carries a charge of 0.25µC.
the radius or the sphere is 150mm what is the field strength at the surface of the sphere?
(surface area of the sphere is 4(pie)R²)

cheers!

Re: Field Strength?

jenko said:

hi guys any help me with this?

A metal Spher, suitably insulated from earth, carries a charge of 0.25µC.
the radius or the sphere is 150mm what is the field strength at the surface of the sphere?
(surface area of the sphere is 4(pie)R2)

cheers!

Click to expand...

There are two basic ways of calculating this:
a) the short way: reasoning and thinking.
b) the long way: mathematically (also thinking, of course)

a) The short way

We all know that the electric field originated by a charge is:

\[E=\frac{1}{4\cdot{\pi}\cdot\epsilon_0}\cdot{\frac{q}{r^2}}\]

This distance r is measured from the center of the charge. In the surface there are two things to take on account: (1) distance to the center is R, the radius of the sphere, and (2) charge contributing to the field is the whole charge of the sphere, so finally we just have to substitute:

\[E=\frac{1}{4\cdot{\pi}\cdot\epsilon_0}\cdot{\frac{q}{r^2}} = \frac{1}{4\cdot{\pi}\cdot\epsilon_0}\cdot{\frac{0.25\cdot{10^{-6}}}{(150\cdot{10^{-3}})^2}}\]

Which gives us:

\[E=100000 C/m\]


b) The long way

First of all we need to calculate the electrical density of the sphere. For that, I assume the charge is homogeneously shared in the sphere, which means the density will be constant, because no other thing is said about it. So, the density will be the total amount of charge divided by the sphere's volume, so first we need to know the sphere's volume and then calculate density:

Total charge: \[Q_{Total}=0.25\mu{C}=0.25\cdot{10^{-6}} C \]
Sphere volume: \[V_{sphere}=\frac{4}{3}\cdot{\pi}\cdot{R^3}=\frac{4}{3}\cdot{\pi}\cdot{{150\cdot{10^{-3}}}^3}=1.414\cdot{10^{-2}} m^3\]

Then electric density in the sphere:
\[\rho = \frac{Q_{Total}}{V_{sphere}}=17.68 {\mu}V/m^3 \]

So the charge inside the sphere measured from the center to a distance r is calculated:
\[q(r) = \int_{0}^{r}\rho\cdot{dr}=\rho\cdot{(r-0)}=\rho\cdot{r}\]

By the other hand, we know that the electric field originated by a charge, to a distance r is:

\[E=\frac{1}{4\cdot{\pi}\cdot\epsilon_0}\cdot{\frac{q}{r^2}}\]

Where in our case, the charge taken on account will be the amount of charge from the center to a distance r, q(r).

Then the electric field:

\[E=\frac{1}{4\cdot{\pi}\cdot\epsilon_0}\cdot{\frac{q(r)}{r^2}}=\frac{1}{4\cdot{\pi}\cdot\epsilon_0}\cdot{\frac{\rho\cdot{r}}{r^2}}=\frac{1}{4\cdot{\pi}\cdot\epsilon_0}\cdot{\frac{\rho}{r}}\]

And, as the surface is the distance R from the center:

\[E=\frac{1}{4\cdot{\pi}\cdot\epsilon_0}\cdot{\frac{\rho}{R}}=\frac{1}{4\cdot{\pi}\cdot\epsilon_0}\cdot{\frac{17.68\cdot{10^{-6}}}{150\cdot{10^{-3}}}}\]

So finally:

\[E\approx{100000} C/m\]


And that's it. Please, somebody correct me if I'm wrong. It's been a long long time since I used to do these kind of things.