• Support PF! Buy your school textbooks, materials and every day products Here!

Relating Electric Field and Voltage in an Insulating Sphere

  • Thread starter NeonJay
  • Start date
  • #1
3
0

Homework Statement



Positive electric charge Q is distributed uniformly throughout the volume of an insulating sphere with radius R. From the expression for E=kQ/r^2 for r>R and E=kQr/R^3 for r<R, find the expression for the electric potential V as a function of r both inside and outside the uniformly charged sphere. Assume that V = 0 at infinity.

k=1/(4*π*ε0)


Homework Equations


[tex]\int_A^B\vec{E}\cdot\mathrm{d}\vec{r} = V(A) - V(B)[/tex]

2.5. Apparent Answer

(kQ/2R)[3-(r^2)/(R^2)]

Apologies ... I'm not very knowledgeable in tex, but that last part would read r-squared over R-squared.

The Attempt at a Solution



I have already obtained V = kQ/r for r>R. This was a simple antiderivative.
I'm getting really caught up on r<R. I've tried integrating from 0 to r, 0 to R, r to R, and backwards versions of the same. Nothing comes to the answer in 2.5 above. I'm really at a loss for how this integral should be done. I can understand most of it except for the 3 in the second expression; where the heck does it come from? I've seen this answer in a couple of different places, but none of them actually explain it.

Any help would be appreciated. Thanks ahead of time.
 

Answers and Replies

  • #2
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,307
999

Homework Statement



Positive electric charge Q is distributed uniformly throughout the volume of an insulating sphere with radius R. From the expression for E=kQ/r^2 for r>R and E=kQr/R^3 for r<R, find the expression for the electric potential V as a function of r both inside and outside the uniformly charged sphere. Assume that V = 0 at infinity.

k=1/(4*π*ε0)


Homework Equations


[tex]\int_A^B\vec{E}\cdot\mathrm{d}\vec{r} = V(A) - V(B)[/tex]

2.5. Apparent Answer

(kQ/2R)[3-(r^2)/(R^2)]

Apologies ... I'm not very knowledgeable in tex, but that last part would read r-squared over R-squared.

The Attempt at a Solution



I have already obtained V = kQ/r for r>R. This was a simple antiderivative.
Hello NeonJay. Welcome to PF !

Extend this to r = R, to get V(R) = kQ/R.

Then use [itex]\displaystyle \int_A^B\vec{E}\cdot\mathrm{d}\vec{r} = V(A) - V(B)[/itex] to get V(r) for r < R . To do this, A = R, B = r .
I'm getting really caught up on r<R. I've tried integrating from 0 to r, 0 to R, r to R, and backwards versions of the same. Nothing comes to the answer in 2.5 above. I'm really at a loss for how this integral should be done. I can understand most of it except for the 3 in the second expression; where the heck does it come from? I've seen this answer in a couple of different places, but none of them actually explain it.

Any help would be appreciated. Thanks ahead of time.
 
  • #3
ehild
Homework Helper
15,478
1,854
The potential is continuous function, and you know it at r=R. V(R)=kQ/R2 . Integrate from r to R:
[tex]V(r)-V(R)=\int_r^R{\frac{kQ}{R^3}rdr}[/tex]
substitute kQ/R for V(R) and solve for V(r).

ehild
 
  • #4
3
0
I think I get it. I was essentially neglecting part of my equation.

Thanks for the help! It really clears a lot of things up.
 
  • #5
ehild
Homework Helper
15,478
1,854
The equation comes from the definition of the electric potential: a function V(r) of position r so as the negative gradient of V is equal to the electric field intensity.

-V(r)=E.


The electric field is conservative so the work done on unit charge between two points A and B does not depend on the path taken.

[tex]\int_A^B{-\nabla V d\vec r}=\int_A^B{\vec Ed\vec r}[/tex]

V dr=dV, the increment of V.

[tex]\int_A^B{-dV}=V(A)-V(B)=\int_A^B{\vec Ed \vec r}[/tex]


ehild
 

Related Threads on Relating Electric Field and Voltage in an Insulating Sphere

  • Last Post
Replies
2
Views
5K
Replies
3
Views
26K
Replies
2
Views
6K
Replies
2
Views
2K
Replies
1
Views
7K
Replies
1
Views
5K
Replies
1
Views
4K
Replies
1
Views
869
Top