How Do You Calculate Electric Potential for Spherical Shells?

In summary: Yes any value of x. Don't equate the potential to zero. Just write the expression first.Edit: Write the potential in terms of the variable.
  • #1
nchin
172
0
Electric potential probelm! HELP PLEASE TEST TMRW!

Two uniformly charged spherical shells of radii R and 2R with identical surface charge densities σ are along the x-axis. The surface of each of the spheres is located a distance R from the origin O
Define the potential at ∞ to be zero.

Along the x-axis are two locations A and B that are R/2 from the Origin (O).

A) What is the potential at A?
B) What is the potential at B?
C) Work required by external foce to move a charge q from A to B?
D) Where between the spheres and along x-axis is the potential the minimum?

Can someone explain this problem to be conceptually on what is going on because i don't uderstand.

this is what i have done so far
A) V = Kq / R ----> q = σ x 4πr2 and distance r = R/2
so VA = 2K(σ x 4π(R)2) / R

B) is the potential at B the same as the potential at A except for the radius is 2R instead of R?
VB = 2K(σ x 4π(2R)2) / R

c)W = qΔV
ΔV = answer to B - answer to A = VA = 2K(σ x 4π(3R)2) / R ?

D)what does this mean? potential the minimum?!

also my teacher did not provide me with a picture or any diagram so I'm kinda visualizing it in my head. HELP ME PLEASE!
 
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  • #2
Check out the attachment. Maybe this is what the diagram may look like.
 

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  • #3
Sunil Simha said:
Check out the attachment. Maybe this is what the diagram may look like.
THANKS!

So would this be correct
Vsphere on the left = Kq / R ----> q = σ x 4πr2 and distance r = 3R - R
so VA = K(σ x 4π(2R)2) / 2R

Vsphere on the right = K(σ x 4π(2R)2) / 4R
K(σ x 4π(R)2) / 4R

im confused on the distance part. is Sphere on the left 2R away from point A and right sphere is 4R away from point A?
 
  • #4
also I am still confused on the part d where it says where the potential is at minimum. how would i solve that part?
 
  • #5
Sorry for the late reply.

In the diagram, A is R/2 from the origin. So its distance from the center of the 2R sphere is 5r/2. And its distance from the R sphere is again 5R/2. So you have to find VA as the sum of the two potentials because of the two spheres.

To find the point where the potential is minimum, consider a point (x,0). Find its potential. Differentiate the expression w.r.t. x and equate to zero. You'll get a value of x where the potential is minimum.
 
  • #6
Sunil Simha said:
Sorry for the late reply.

In the diagram, A is R/2 from the origin. So its distance from the center of the 2R sphere is 5r/2. And its distance from the R sphere is again 5R/2. So you have to find VA as the sum of the two potentials because of the two spheres.

To find the point where the potential is minimum, consider a point (x,0). Find its potential. Differentiate the expression w.r.t. x and equate to zero. You'll get a value of x where the potential is minimum.


wait hows it 5R from the origin? The center of the 2R sphere is 3R from the origin? and the center of R sphere is 2R from the origin?
 
  • #7
nchin said:
wait hows it 5R from the origin? The center of the 2R sphere is 3R from the origin? and the center of R sphere is 2R from the origin?

How's what 5R from the origin?
 
  • #8
Sunil Simha said:
How's what 5R from the origin?


You said the distance from the center of the 2R sphere is 5r/2. And its distance from the R sphere is again 5R/2. i don't understand how its 5R from the point of the sphere to the origin (0,0) on the graph
 
  • #9
also what does the expression w.r.t. mean
 
  • #10
I'm really sorry for my poor use of grammar there. Let me explain again. From the center of the sphere of radius 2R, the distance of A is 5R/2 right? Also from the sphere of radius R, the distance of A is 5R/2. So now find the potential at A because of the two spheres.

w.r.t means with respect to. Differentiate the expression for potential at (x,0) w.r.t. x and equate it to zero to obtain the x where the potential in minimal.
 
  • #11
Sunil Simha said:
I'm really sorry for my poor use of grammar there. Let me explain again. From the center of the sphere of radius 2R, the distance of A is 5R/2 right? Also from the sphere of radius R, the distance of A is 5R/2. So now find the potential at A because of the two spheres.

w.r.t means with respect to. Differentiate the expression for potential at (x,0) w.r.t. x and equate it to zero to obtain the x where the potential in minimal.

i see. so for point b, the two spheres would have different distances right?
 
  • #12
nchin said:
i see. so for point b, the two spheres would have different distances right?

Yes. Point B would have different set of distances from the centers of the spheres. Find the potential there too.

I hope you have understood how to obtain the point where the potential is minimum.
 
  • #13
Sunil Simha said:
Yes. Point B would have different set of distances from the centers of the spheres. Find the potential there too.

I hope you have understood how to obtain the point where the potential is minimum.


i don't really understand the minimum potential part.
 
  • #14
nchin said:
i don't really understand the minimum potential part.

Ok. First choose an arbitrary point (x.0). What is the net potential at this point?
 
  • #15
Sunil Simha said:
Ok. First choose an arbitrary point (x.0). What is the net potential at this point?

as in we can choose any x value? we equate the potential to zero?
 
  • #16
nchin said:
as in we can choose any x value? we equate the potential to zero?

Yes any value of x. Don't equate the potential to zero. Just write the expression first.

Edit: Write the potential in terms of the variable x
 
  • #17
Sunil Simha said:
Yes any value of x. Don't equate the potential to zero. Just write the expression first.

Edit: Write the potential in terms of the variable x

v = kq/x
 
  • #18
No! No!

What is the potential at (x,0) due to the two spheres? How did you find potential at A? Do the same thing but choose the point(x,0)
 
  • #19
Sunil Simha said:
No! No!

What is the potential at (x,0) due to the two spheres? How did you find potential at A? Do the same thing but choose the point(x,0)

the potential due to the two spheres at point A is the sum of the potential of the two sphere so kq/r + kq/r with r = the distance right?

so since we're using the point x,0 then shouldn't it be kq / x?
 
  • #20
nchin said:
the potential due to the two spheres is kq/r + kq/r with r = the distance right?

You've got the concept but the equation is still a bit wrong. Firstly the charges on both the spheres is not the same right? Then the distance of the point (x,0) from both the spheres in not the same, is it?
 
  • #21
Sunil Simha said:
You've got the concept but the equation is still a bit wrong. Firstly the charges on both the spheres is not the same right? Then the distance of the point (x,0) from both the spheres in not the same, is it?

identical surface charge density = same charge right? q = σ x A ---> A = 4πr2. but the only difference are the radius.

(kq/3R - x) + (kq/2R - x)?
 
  • #22
nchin said:
identical surface charge density = same charge right? q = σ x A ---> A = 4πr2. but the only difference are the radius.

kq/3R - x and kq/2R - x?

Since their radius is different, won't the charges be different because the areas are not the same.

In the potential equation I recommend you take the the distances as |3R-x| and |2R-x|. Then you can take cases to find the potential when x>2R, -3R<x<2R and x<-3R.

Once you get the potential, for each interval of x ( as I have divided in my previous sentence), differentiate the expression w.r.t. x and equate that to zero. You should get some values of x corresponding to dV/dx = 0. Substitute them in the original equation and find the potential. Choose the minimum potential and the answer to the question would be the corresponding value of x.
 
  • #23
Sunil Simha said:
Since their radius is different, won't the charges be different because the areas are not the same.

In the potential equation I recommend you take the the distances as |3R-x| and |2R-x|. Then you can take cases to find the potential when x>2R, -3R<x<2R and x<-3R.

Once you get the potential, for each interval of x ( as I have divided in my previous sentence), differentiate the expression w.r.t. x and equate that to zero. You should get some values of x corresponding to dV/dx = 0. Substitute them in the original equation and find the potential. Choose the minimum potential and the answer to the question would be the corresponding value of x.

so i take the derivative of kq/3R -x then set that derivative equal to 0??
 
  • #24
nchin said:
so i take the derivative of kq/3R -x then set that derivative equal to 0??

You take the derivative of the sum (i.e. the net potential) and then equate it to zero.
 
  • #25
Sunil Simha said:
You take the derivative of the sum (i.e. the net potential) and then equate it to zero.

that derivative seems tough. do i use the quotient rule for that?
 
  • #26
nchin said:
that derivative seems tough. do i use the quotient rule for that?

Yes you have to use quotient rule and the derivative is not tough. You should be able to do it within a couple of steps. (Remember k and Q for a sphere is constant, so you're simply finding the derivative of 1/distance.)
 
  • #27
Sunil Simha said:
Yes you have to use quotient rule and the derivative is not tough. You should be able to do it within a couple of steps. (Remember k and Q for a sphere is constant, so you're simply finding the derivative of 1/distance.)

ooh yea they are constant. thanks for your help!
 
  • #28
You're welcome.:smile:

Best of luck for tomorrow.
 

1. What is an electric potential problem?

An electric potential problem involves the calculation of the electric potential at a given point in space, caused by a distribution of charges.

2. How is electric potential different from electric field?

Electric potential is a scalar quantity that measures the potential energy of a unit charge at a given point in space, while electric field is a vector quantity that measures the force experienced by a unit charge at a given point in space.

3. What is the equation for electric potential?

The equation for electric potential is V = kQ/r, where V is the electric potential, k is the Coulomb's constant, Q is the charge of the source, and r is the distance between the source and the point of interest.

4. How do you solve an electric potential problem?

To solve an electric potential problem, you first need to identify the charges and their positions. Then, you can use the equation V = kQ/r to calculate the potential at the point of interest. Make sure to pay attention to the signs of the charges and their distances.

5. What are some real-life applications of electric potential problems?

Electric potential problems are important in understanding the behavior of electric circuits, the design of electronic devices, and the functioning of the nervous system in living organisms. They are also used in electrostatics experiments and in the treatment of certain medical conditions.

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