Question

(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.)

Answer 1

@Michele_Laino

Answer 2

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

Answer 3

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

Answer 4

inside the sphere there will be constant potential, right?

Answer 5

yes! since inside the sphere there is no electric field

Answer 6

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

Answer 7

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

Answer 8

what about b) ?

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

that's really awesome

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

ok now i get what is inside the integral

Answer 17

However idk where did you get that density

Answer 18

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

Answer 19

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

Answer 20

mm I see,
did you study physics i guess

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

and of course even more differential equations ..

Answer 27

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 \(t\)

Answer 28

variables*

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

gradient is a directional derivative, right ?

Answer 33

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

Answer 34

damn This is really nice sometimes math could be outstanding

Answer 35

of course! :)