Electric Potential V inside nonconducting sphere with cavity

In summary: It's not standard to write V_R_2, but it would have been clearer if the author had shown the evaluation.
  • #1
WK95
139
1

Homework Statement


A nonconducting sphere of radius r_2 contains a concentric spherical cavity of radius r_1. The material between r_1 and r_2 carries a uniform charge density rho_E(C/m^3). Determine the electric potential V, relative to V=0 at r= infinity, as a function of the distance r from the center for r_1 < r < r_2

Homework Equations

The Attempt at a Solution


http://i.imgur.com/bOMD7ej.png

The link is just a screenshot of the answer and steps given in the solution manual. I have similar work but I am confused about why the solution involves integrating from r_2 to r. How is that the potential relative to V=0 at r=infinity?

And what happened to V_R_2? Since it disappears later on, I assume it's because it's equal to 0V but how come?
 

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  • #2
WK95 said:
The link is just a screenshot of the answer and steps given in the solution manual. I have similar work but I am confused about why the solution involves integrating from r_2 to r. How is that the potential relative to V=0 at r=infinity?

And what happened to V_R_2? Since it disappears later on, I assume it's because it's equal to 0V but how come?

You know that the negative gradient of the potential is equal to the electric field. Getting the potential function, find the electric field at the different domains. You get the potential at a point r if integrating between r and a point where you know the potential. Do you know the potential at r2, the radius of the outer surface of the shell? What is the electric field outside the shell?
 
  • #3
WK95 said:
the solution involves integrating from r_2 to r. How is that the potential relative to V=0 at r=infinity?
It isn't. The potential at r is that at r2 minus the integral from r to r2. The potential at infinity is the same as it would be at r if the body were uncharged, or if r1 = r = r2.
WK95 said:
And what happened to V_R_2?
It was evaluated as (something like) ##\frac{4\pi}3\left(r_2^2-\frac{r_1^3}{r_2}\right)\frac{\rho}{4\pi\epsilon_0}## and combined with the result of the integral. Write out the steps and you'll see the result is correct.
 

FAQ: Electric Potential V inside nonconducting sphere with cavity

What is electric potential V inside a nonconducting sphere with a cavity?

The electric potential V inside a nonconducting sphere with a cavity is the measure of the electric potential energy per unit charge at any given point inside the sphere. It is influenced by the distribution of charges within the sphere and the distance from the center of the sphere to the point of interest.

How is the electric potential V inside a nonconducting sphere with a cavity calculated?

The electric potential V inside a nonconducting sphere with a cavity can be calculated using the formula V = kQ/R, where k is the Coulomb's constant, Q is the charge inside the cavity, and R is the distance from the center of the sphere to the point of interest.

What is the significance of a cavity in a nonconducting sphere?

The cavity in a nonconducting sphere allows for a varying distribution of charges within the sphere, which affects the electric potential V inside the sphere. It also allows for different electric field strengths at different points inside the sphere, which can have practical applications in various fields of science and technology.

Can the electric potential V inside a nonconducting sphere with a cavity be negative?

Yes, the electric potential V inside a nonconducting sphere with a cavity can be negative. This means that the electric field is directed towards the center of the cavity, and the electric potential energy decreases as the distance from the center decreases.

How does the electric potential V inside a nonconducting sphere with a cavity change with the size of the cavity?

The electric potential V inside a nonconducting sphere with a cavity is directly proportional to the size of the cavity. This means that as the size of the cavity increases, the electric potential inside the sphere also increases. However, the exact relationship between the two varies depending on the specific conditions and distribution of charges within the sphere.

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