(a) How much charge is on the surface of an isolated
spherical conductor that has a radius 10cm and is charged to 2KV
(b) What is the electrostatic potential energy of this conductor?
(Assume the potential is zero far from the sphere.)

- rock_mit182

- jamiebookeater

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- rock_mit182

- rock_mit182

I guess if the radius is infinite the potential is 0.. but im stuck from there

- Michele_Laino

the relationship between the potential \(V\) on the surface of our sphere, and its charge is:
\[V = \frac{1}{{4\pi {\varepsilon _0}}}\frac{Q}{a}\]
where \(a\) is the radius of our sphere

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## More answers

- rock_mit182

inside the sphere there will be constant potential, right?

- Michele_Laino

yes! since inside the sphere there is no electric field

- Michele_Laino

It is supposed that the electric charge is only on the surface of the sphere

- rock_mit182

ok i got the first part 2* 10^3 V (4*pi*epsilon_o)(.1) m =Q

- rock_mit182

what about b) ?

- Michele_Laino

for part (b), we have to consider the energy trapped inside the electrostatic field outside the sphere

- rock_mit182

that's why faraday's cage is a success ?

- Michele_Laino

yes! inside a metallic box there is no electric field when, of course, outside that metallic box there is one or more than one electric field

- rock_mit182

that's really awesome

- Michele_Laino

yes! that behaviour it is due to the fact that the electrons of the metallic box will redistribute themselves such that they create an electric field which cancel the external electric field

- Michele_Laino

for part (b), we have to compute this integral:
\[\Large W = \int_a^{ + \infty } {\frac{1}{{8\pi }}} \frac{{{Q^2}}}{{{r^4}}}4\pi {r^2}dr\]
where I have used the \(CGS\) system

- Michele_Laino

since I assign a density energy of \[\frac{{{E^2}}}{{8\pi }}\] in the outer space

- rock_mit182

ok now i get what is inside the integral

- rock_mit182

However idk where did you get that density

- Michele_Laino

of course we can show why we have that energy density( \(energy/Volume\) ), nevertheless, when I make computations similar to your exercise I simply remember the formula of that density

- Michele_Laino

I remark that I have used \(CGS\) system, since when I was at university I have studied electromagnetism using that system

- rock_mit182

mm I see,
did you study physics i guess

- Michele_Laino

yes! I have a degree in Physics
the electric field, written in \(CGS\) system is:
\[E = \frac{Q}{{{r^2}}}\]

- rock_mit182

well I have another question, maybe is unrealated to this topic, Is there such thing as magnetic potential or electromagnetic potential ?

- Michele_Laino

more precisely there is a vector potential \(A\) for magnetic field \(B\), such that the subsequent vector equation holds:
\[{\mathbf{B}} = {\text{rot }}{\mathbf{{\rm A}}}\]

- Michele_Laino

it is a partial differential equation, and \(rot\) stands for the \(rotor\) operator applied to the vector \(A\)

- rock_mit182

i guess I have to learn multivariable calculus to understand such concepts...

- rock_mit182

and of course even more differential equations ..

- Michele_Laino

yes! correct! Since the vector \(A\) is a function of the spatial variable \(x,y,z\) and, for time-dependent fields, it is also a function of time \(t\)

- Michele_Laino

variables*

- rock_mit182

well dude you have been very helpful, now i have to study capacitors if a qeustion comes along I'll looking for you. thanks for helping me out

- Michele_Laino

thanks! :)
Finally, we can write this analogy for both electric and magnetic field:
Electric field:
\[{\mathbf{E}} = grad\varphi \]
and, magnetic field:
\[{\mathbf{B}} = {\text{rot }}{\mathbf{{\rm A}}}\]
the functions \(\phi\) and \(A\) play analogue role

- Michele_Laino

\(grad\) stands for \(gradient\) operator applied to function \(\phi\)

- rock_mit182

gradient is a directional derivative, right ?

- Michele_Laino

yes! In cartesian coordinates we have this formula:
\[{\text{grad}}\varphi = \left( {\frac{{\partial \varphi }}{{\partial x}},\frac{{\partial \varphi }}{{\partial y}},\frac{{\partial \varphi }}{{\partial z}}} \right)\]
it is a vector

- rock_mit182

damn This is really nice sometimes math could be outstanding

- Michele_Laino

of course! :)

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