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rock_mit182
 one year ago
(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
 one year ago
(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.)

This Question is Closed

rock_mit182
 one year ago
Best ResponseYou've already chosen the best response.0I guess if the radius is infinite the potential is 0.. but im stuck from there

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1the 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

rock_mit182
 one year ago
Best ResponseYou've already chosen the best response.0inside the sphere there will be constant potential, right?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes! since inside the sphere there is no electric field

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1It is supposed that the electric charge is only on the surface of the sphere

rock_mit182
 one year ago
Best ResponseYou've already chosen the best response.0ok i got the first part 2* 10^3 V (4*pi*epsilon_o)(.1) m =Q

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1for part (b), we have to consider the energy trapped inside the electrostatic field outside the sphere

rock_mit182
 one year ago
Best ResponseYou've already chosen the best response.0that's why faraday's cage is a success ?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes! 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
 one year ago
Best ResponseYou've already chosen the best response.0that's really awesome

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes! 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
 one year ago
Best ResponseYou've already chosen the best response.1for 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
 one year ago
Best ResponseYou've already chosen the best response.1since I assign a density energy of \[\frac{{{E^2}}}{{8\pi }}\] in the outer space

rock_mit182
 one year ago
Best ResponseYou've already chosen the best response.0ok now i get what is inside the integral

rock_mit182
 one year ago
Best ResponseYou've already chosen the best response.0However idk where did you get that density

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1of 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
 one year ago
Best ResponseYou've already chosen the best response.1I remark that I have used \(CGS\) system, since when I was at university I have studied electromagnetism using that system

rock_mit182
 one year ago
Best ResponseYou've already chosen the best response.0mm I see, did you study physics i guess

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes! I have a degree in Physics the electric field, written in \(CGS\) system is: \[E = \frac{Q}{{{r^2}}}\]

rock_mit182
 one year ago
Best ResponseYou've already chosen the best response.0well I have another question, maybe is unrealated to this topic, Is there such thing as magnetic potential or electromagnetic potential ?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1more 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
 one year ago
Best ResponseYou've already chosen the best response.1it is a partial differential equation, and \(rot\) stands for the \(rotor\) operator applied to the vector \(A\)

rock_mit182
 one year ago
Best ResponseYou've already chosen the best response.0i guess I have to learn multivariable calculus to understand such concepts...

rock_mit182
 one year ago
Best ResponseYou've already chosen the best response.0and of course even more differential equations ..

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes! correct! Since the vector \(A\) is a function of the spatial variable \(x,y,z\) and, for timedependent fields, it is also a function of time \(t\)

rock_mit182
 one year ago
Best ResponseYou've already chosen the best response.0well 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
 one year ago
Best ResponseYou've already chosen the best response.1thanks! :) 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
 one year ago
Best ResponseYou've already chosen the best response.1\(grad\) stands for \(gradient\) operator applied to function \(\phi\)

rock_mit182
 one year ago
Best ResponseYou've already chosen the best response.0gradient is a directional derivative, right ?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes! 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
 one year ago
Best ResponseYou've already chosen the best response.0damn This is really nice sometimes math could be outstanding
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