rock_mit182
  • rock_mit182
(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.)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
rock_mit182
  • rock_mit182
@Michele_Laino
rock_mit182
  • rock_mit182
I guess if the radius is infinite the potential is 0.. but im stuck from there
Michele_Laino
  • 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

rock_mit182
  • rock_mit182
inside the sphere there will be constant potential, right?
Michele_Laino
  • Michele_Laino
yes! since inside the sphere there is no electric field
Michele_Laino
  • Michele_Laino
It is supposed that the electric charge is only on the surface of the sphere
rock_mit182
  • rock_mit182
ok i got the first part 2* 10^3 V (4*pi*epsilon_o)(.1) m =Q
rock_mit182
  • rock_mit182
what about b) ?
Michele_Laino
  • Michele_Laino
for part (b), we have to consider the energy trapped inside the electrostatic field outside the sphere
rock_mit182
  • rock_mit182
that's why faraday's cage is a success ?
Michele_Laino
  • 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
  • rock_mit182
that's really awesome
Michele_Laino
  • 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
  • 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
  • Michele_Laino
since I assign a density energy of \[\frac{{{E^2}}}{{8\pi }}\] in the outer space
rock_mit182
  • rock_mit182
ok now i get what is inside the integral
rock_mit182
  • rock_mit182
However idk where did you get that density
Michele_Laino
  • 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
  • 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
  • rock_mit182
mm I see, did you study physics i guess
Michele_Laino
  • Michele_Laino
yes! I have a degree in Physics the electric field, written in \(CGS\) system is: \[E = \frac{Q}{{{r^2}}}\]
rock_mit182
  • 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
  • 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
  • Michele_Laino
it is a partial differential equation, and \(rot\) stands for the \(rotor\) operator applied to the vector \(A\)
rock_mit182
  • rock_mit182
i guess I have to learn multivariable calculus to understand such concepts...
rock_mit182
  • rock_mit182
and of course even more differential equations ..
Michele_Laino
  • 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
  • Michele_Laino
variables*
rock_mit182
  • 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
  • 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
  • Michele_Laino
\(grad\) stands for \(gradient\) operator applied to function \(\phi\)
rock_mit182
  • rock_mit182
gradient is a directional derivative, right ?
Michele_Laino
  • 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
  • rock_mit182
damn This is really nice sometimes math could be outstanding
Michele_Laino
  • Michele_Laino
of course! :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.