Finding electric potential of a sphere with a given volume charge density

In summary, the conversation discusses determining the electric potential V for a point inside a sphere with radius r0 and volume charge density Pe= Po (1- (r^2/r0^2)). The equations used include E= Qencl/ eps0*A= kQ/r^2, Qencl= int(Pe*4pir^2) dr, and -int(E)= V. The solution involves finding Qencl and evaluating it from 0 to r, resulting in E= Po/eps0 ((r/3)- r^3/ 5r0^2) and V= p0/eps0 ( (r^2/ 6) - (r^4/20r0^
  • #1
citra
6
0

Homework Statement


the volume charge density Pe= Po (1- (r^2/r0^2))
the sphere has a radius of r0
and r is measured from the center
for a point P inside the sphere (r<r0), determine the electric potential V if V=0 at infinity

Homework Equations


E= Qencl/ eps0*A= kQ/r^2
Qencl= int(Pe*4pir^2) dr
-int(E)= V

The Attempt at a Solution


first i found Qencl and evaluated it from 0 to r.
Qencl= 4pi*p0 (r^3/ 3 - r^5/5r0^2)
Then E= Po/eps0 ((r/3)- r^3/ 5r0^2)
then the -integral of E= V= p0/eps0 ( (r^2/ 6) - (r^4/20r0^2) - (7r02/60))
but this is the worng answer..any help?
 
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  • #2
Hi citra, welcome to PF.
You have given volume charge density. But it does not appear correct. Can you check it.
 
  • #3


Your approach is correct, but there may be a mistake in your calculations. Let's go through it step by step:

1. Finding Qencl:

Qencl = int(Pe*4pi*r^2) dr

= 4pi * int(Po (1- (r^2/r0^2)) * r^2) dr

= 4pi * Po * (r^3/3 - r^5/5r0^2) + C

Note that you are missing the constant C in your calculation. This is the integration constant and it is important to include it in your final answer.

2. Finding E:

E = k * Qencl / r^2

= k * 4pi * Po * (r^3/3 - r^5/5r0^2) / r^2

= 4pi * Po * (r/3 - r^3/5r0^2)

Note that you are missing the constant k in your calculation. This is the Coulomb's constant and it is important to include it in your final answer.

3. Finding V:

V = -int(E) = -int(4pi * Po * (r/3 - r^3/5r0^2)) dr

= -4pi * Po * (r^2/6 - r^4/20r0^2) + C

= -4pi * Po * (r^2/6 - r^4/20r0^2) + V0

Note that you are missing the constant V0 in your calculation. This is the potential at r=0 and it is important to include it in your final answer.

Therefore, the final expression for the electric potential is:

V = -4pi * Po * (r^2/6 - r^4/20r0^2) + V0

where V0 is the potential at r=0 and can be found using the given information that V=0 at infinity. Therefore, V0 = 0.

Hence, the final expression for the electric potential at a point P inside the sphere is:

V = -4pi * Po * (r^2/6 - r^4/20r0^2)

I hope this helps!
 

1. How do you calculate the electric potential of a sphere with a given volume charge density?

To calculate the electric potential of a sphere with a given volume charge density, you can use the formula V = kQ/R, where V is the electric potential, k is the Coulomb's constant, Q is the total charge of the sphere, and R is the radius of the sphere.

2. What is the unit for electric potential?

The unit for electric potential is volts (V).

3. Can the electric potential of a sphere with a given volume charge density be negative?

Yes, the electric potential of a sphere can be negative if the charge is negative. This means that the electric potential at a point in space around the sphere is negative, indicating that it is at a lower energy level compared to a point with zero electric potential.

4. What is the relationship between electric potential and electric field?

Electric potential is directly related to electric field. The electric field is the gradient of the electric potential, meaning that the direction of the electric field is in the direction of decreasing electric potential. In other words, the electric field points towards areas of lower electric potential.

5. How does the electric potential change as you move away from the sphere?

The electric potential decreases as you move away from the sphere. This is because the electric field strength decreases with distance, resulting in a decrease in electric potential. This decrease follows an inverse relationship, so the farther you move from the sphere, the lower the electric potential becomes.

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