(General question)Area charge density and electric potential

In summary, the electric potential of a spherical shell of charge Q with radius R is V=\frac{Q}{4\pi\epsilon_0 R} and the electric field outside the shell is \vec E=\frac{Q}{4\pi\epsilon_0 r^2}\hat r.
  • #1
prodo123
17
4

Homework Statement


Sorry, the post isn't about a single homework problem but rather something that I keep getting confused on. It's about calculating the electric potential of a spherical shell of uniform charge in two different ways.

Homework Equations


##\Delta V=\int_a^b -\vec E\cdot d\vec r##
##V=\frac{1}{4\pi\epsilon_0}\int\frac{dq}{r}##
Gauss's Law

The Attempt at a Solution


Taking ##V=0## at ##r\to\infty##, the electric potential of a spherical shell of charge ##Q## with radius ##R## should be:

##V=\int^R_\infty -\vec E\cdot d\vec r##

Using Gauss's Law, the electric field outside the shell is ##\vec E = \frac{Q}{4\pi \epsilon_0 r^2}\hat r##. This gives ##V=\frac{Q}{4\pi\epsilon_0}(\frac{1}{R}-\frac{1}{\infty})=\frac{Q}{4\pi\epsilon_0 R}##, which is what we expect.
The charge area density for the shell is ##\sigma=\frac{Q}{4\pi R^2}=\frac{q}{A}##. Having A be the surface area for a sphere, the function for the charge of a spherical shell of radius r is:

##q(r)=\sigma A = \frac{Q}{4\pi R^2}(4\pi r^2)=Q\frac{r^2}{R^2}##

The infinitesimal charge is therefore ##dq=\frac{2Q}{R^2}r dr##.

Since electric potentials for a distribution of charge is found by ##V=\frac{1}{4\pi\epsilon_0}\int\frac{dq}{r}##, the ##r## variables cancel out and the integral reduces to ##V=\frac{1}{4\pi\epsilon_0}\int^R_0\frac{2Q}{R^2}dr##.

Evaluating the integral gives ##V=\frac{Q}{2\pi\epsilon_0 R}##, which is not what's expected. There's a factor of 2 somewhere that's throwing off the answers. Where am I going wrong?
 
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  • #2
prodo123 said:
a spherical shell of uniform charge
How thick is the shell ? You integrate from 0 to R but is that really where the charge resides ?
 
  • #3
BvU said:
How thick is the shell ? You integrate from 0 to R but is that really where the charge resides ?
##A_{\text{sphere}}=R^2\iint sin(\theta)d\theta d\phi## where ##\pi \ge \theta \ge 0## and ##2\pi\ge\phi\ge 0## for an infinitely thin spherical shell
Leaving ##\phi## as a variable gives the following:
##A=2R^2\int d\phi##
##dA = 2R^2 d\phi##
##q=\sigma A##
##dq = \sigma dA = 2\sigma R^2 d\phi##
##V=\frac{1}{4\pi\epsilon_0}\int^{2\pi} _0 (\frac{1}{R})2\sigma R^2 d\phi##
##V=\frac{\sigma R}{\epsilon_0} = \frac{\frac{Q}{4\pi R^2}R}{\epsilon_0}##
##V=\frac{Q}{4\pi\epsilon_0 R}##
Right...thanks!
 

FAQ: (General question)Area charge density and electric potential

1. What is the definition of area charge density?

Area charge density, denoted by the symbol ρ, is a measure of the amount of electric charge per unit area. It is typically expressed in units of coulombs per square meter (C/m²).

2. How is area charge density related to electric potential?

The electric potential at a point is directly proportional to the area charge density at that point. This means that as the area charge density increases, the electric potential also increases, and vice versa.

3. Can the area charge density be negative?

Yes, the area charge density can be negative if the charge distribution at a point is negative. This indicates an excess of negative charges compared to positive charges at that point.

4. How is area charge density calculated?

Area charge density can be calculated by dividing the total charge present in a given area by the area itself. Mathematically, it can be expressed as ρ = Q/A, where Q is the total charge and A is the area.

5. What are some real-life applications of area charge density and electric potential?

Area charge density and electric potential have various applications in everyday life, such as determining the capacitance of a capacitor, analyzing the behavior of conductors and insulators, and designing electronic circuits and devices.

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