Electric potential in different cases

[quietrain]

ok, the electric field E = - dV / dx, where V is the potential. but what is exactly is x?

because i am quite confused by E and V

why is it that the electric field E is = 0 IN a conducting charged sphere of radius R, AND V = k (q/R) ,

WHEREAS, the electric field E is = 0 IN a long positively charged cylinder with uniform charge per unit length along its axis , AND V = 0 here.

so why is there a value for V in a conducting charged sphere and not in a long positively charged cylinder with uniform charge per unit length?

so is there anything special that one should look out for to determine whether is the potential V inside anything 0 or something? how do we know?

thanks a lot!

 

[zachzach]

ok, the electric field E = - dV / dx, where V is the potential. but what is exactly is x?

because i am quite confused by E and V

why is it that the electric field E is = 0 IN a conducting charged sphere of radius R, AND V = k (q/R) ,

WHEREAS, the electric field E is = 0 IN a long positively charged cylinder with uniform charge per unit length along its axis , AND V = 0 here.

so why is there a value for V in a conducting charged sphere and not in a long positively charged cylinder with uniform charge per unit length?

so is there anything special that one should look out for to determine whether is the potential V inside anything 0 or something? how do we know?

thanks a lot!

E = - dV / dx ---> What does this equation mean? It means the electric field is equal to the rate of change of the potential with respect to distance. In this case it would calculate the electric field in the x direction. You could similarly compute E = -dV/dy to compute the E-field in the y direction. For a spherically symmetric system you can use E = -dV/dr to compute the E-field in the radial direction. As for the two examples, what do you know about functions whose derivatives = 0?

 

[quietrain]

if E = 0 that means there will be no change in the potential over the distance? which means dV/dx =0?

so since both examples have E = 0, why does the sphere has a expression for V , which is kQ/R, whereas the long cylinder with uniform charge density has just V = 0? ,

since E = -dV/dr for the sphere, and E =0, shouldn't -dV/dr = 0 , so V is just any constant?

oh...

so now then why does the long cylinder has exactly V =0 and not just any constant?

 

[zachzach]

if E = 0 that means there will be no change in the potential over the distance? which means dV/dx =0?

so since both examples have E = 0, why does the sphere has a expression for V , which is kQ/R, whereas the long cylinder with uniform charge density has just V = 0? ,

since E = -dV/dr for the sphere, and E =0, shouldn't -dV/dr = 0 , so V is just any constant?

oh...

so now then why does the long cylinder has exactly V =0 and not just any constant?

If the derivative of a function is zero that means that the function is a constant. So any constant function has a derivative of zero. This does NOT mean that a function whose derivative is zero is any constant function. The constant is going to depend on the parameters in the problem. In this case the geometry and the charge distribution. Also, the expression V=0 is a constant as much as V = kQ/R is. Try solving for V in both cases and see what you get. Hint: Does V have to be continuous at the boundary? If so why? If not why not?

 

[quietrain]

oh, so you mean that because the charged conducting sphere has charges distributed all over in the sphere, it's V will depend on the geometry and so it has V = kQ/r?

but on the other hand the linear charge density line charge cylinder, has charges lined up in a row, so the equation has continuous distribution of charges ,so V = 0? but why?

 

[zachzach]

oh, so you mean that because the charged conducting sphere has charges distributed all over in the sphere, it's V will depend on the geometry and so it has V = kQ/r?

but on the other hand the linear charge density line charge cylinder, has charges lined up in a row, so the equation has continuous distribution of charges ,so V = 0? but why?

No. I am saying that the potential of any object is dependent on the geometry and the charge distribution. The potential of the cylinder is dependent on the geometry and charge distribution just as much as the sphere is. Once you figure out what V is in both cases using math you will pop out the answers. And because the potential of both objects does not change inside them (it is a constant potential) the derivative is 0 and thus the E field is as well. Do you know how to find V for both objects? Why is V = 0? Basically this is just what the charge distributions and geometry give us once V is computed.

 

[ApexOfDE]

in conductors, all charges will concentrate on the surface. That's why there is no field inside a conducting sphere. However, there is field outside the sphere. Calculate it, then by taking its integrals (from r to inf.) to find potential inside sphere.

p/s: I am confused with cylinder. Can you quote a whole problem? Sry for this :(

 

[quietrain]

in conductors, all charges will concentrate on the surface. That's why there is no field inside a conducting sphere. However, there is field outside the sphere. Calculate it, then by taking its integrals (from r to inf.) to find potential inside sphere.

p/s: I am confused with cylinder. Can you quote a whole problem? Sry for this :(

integrate -Edr = dV ?
but why do we have to find the integrals from r to inf which is OUTSIDE the sphere to find the potential INSIDE a sphere?

anyway, E = kQ/ r², but how do we integrate? since Q is not constant? i can't use constant surface charge density right? since Q/4(pi)r² would be constant, i would have no r terms in my electric field equation to integrate? or am i doing something wrong?

about the cylinder, its basically an infinite long charged cylinder with constant linear charge density.

so basically my question is: why for a conducting sphere of radius R, we have E = 0, V = kQ/r inside the sphere (i.e. r<R) ,
On the other hand, for an infinite long charged cylinder with radius R, inside the cylinder, we have E = 0 , BUT V= 0 here, compared to the V term in the sphere.

 

[quietrain]

No. I am saying that the potential of any object is dependent on the geometry and the charge distribution. The potential of the cylinder is dependent on the geometry and charge distribution just as much as the sphere is. Once you figure out what V is in both cases using math you will pop out the answers. And because the potential of both objects does not change inside them (it is a constant potential) the derivative is 0 and thus the E field is as well. Do you know how to find V for both objects? Why is V = 0? Basically this is just what the charge distributions and geometry give us once V is computed.

V = kQ/r for the conducting sphere? so we get this by taking E r = V?

but i am not sure about the cylinder's potential V. my notes says its V = (lamda) / 2(pi)(e₀)r , so E = -dV/dr

but these formulas are for r > Radius of cylinder.
for r < Radius, i.e in the cylidner itself, why is E = 0 and V =0 ?

so you pointed out that it was due to geometry and charge distribution. so the charge distributtion of the conducting sphere is all on the surface, but for the cylinder it is spread out within the cylinder. so is that why V = 0 in the cylinder and V = some term for the sphere?

but if i sub r < Radius into the potential equation V = (lamda) / 2(pi)(e₀)r, i don't get 0 ?

 

[quietrain]

oh i think i am getting it. the V for a sphere is given by V = kQ/r. so the r term takes the value of Radius R for the conducting sphere since all the charges are on the surface, so all of the charges have distance R.

but what about the long cylinder linear charge density. its V is given by V= (lamda)/2(pi)(e) * (ln R/r) so if we sub r < R, i.e inside the sphere, wouldn't the ln term tend to infinity? since R/r will tend to infinity with decreasing r ?

so how do we ascertain that V = 0 in this case?

 

[ApexOfDE]

in sphere, integral intervals [r inf] ==> [r R] + [R inf]. Because there is no field inside sphere, only the 2nd integrals survives.

 

[zachzach]

oh i think i am getting it. the V for a sphere is given by V = kQ/r. so the r term takes the value of Radius R for the conducting sphere since all the charges are on the surface, so all of the charges have distance R.

The equation V = KQ/r is the potential (only for outside the sphere) with respect to whatever distance you are from the origin. r is just your position from the origin. On the boundary (surface of sphere) V = KQ/R since the distance from the origin is R (obviously). But you want V INSIDE the sphere. You know that E = 0 everywhere inside from Gauss's Law and there being no charge inside the sphere. So you also know that V is a constant inside the sphere since E = -dv/dr = 0 and funcions whose derivatives are 0 are constants. How do you find the constant? V = KQ/R on the boundary and let's just guess that V = 0 inside. What would that mean for E on the surface? At that point V would be discontinuous (going form 0 --> KQ/R at the boundary) and thus the derivative would not exist at this point. so E = -dv/dr would be undefined! In order for the derivative to exist V has to be continuous at the boundary and since we know V = KQ/R at the boundary and we know inside the sphere V is the same everywhere then V must = KQ/R everywhere inside the sphere.

but what about the long cylinder linear charge density. its V is given by V= (lamda)/2(pi)(e) * (ln R/r) so if we sub r < R, i.e inside the sphere, wouldn't the ln term tend to infinity? since R/r will tend to infinity with decreasing r ?

so how do we ascertain that V = 0 in this case?

Now see if you can see why V = 0 inside the cylinder. Hint: use your equation V= (lamda)/2(pi)(e) * (ln R/r).

 

[quietrain]

oh! so by similar reasoning, to have continuous function of potential, we have lnR/R = ln 1 = 0. so since the potential on the surface of the cylinder is 0, everywhere in there has to be 0?

i see.. thanks!