Electric field at a point inside a non-uniformly charged sphere

AI Thread Summary
The discussion centers on calculating the electric field inside a non-uniformly charged sphere using Gauss's law. The initial approach involves taking derivatives of charge with respect to area, leading to an incorrect expression for the electric field. Participants emphasize the importance of integrating to find charge before applying Gauss's law, rather than differentiating. There is a specific concern about discrepancies when substituting values for a uniformly charged sphere, which suggests a misunderstanding in the application of the law. Ultimately, the conversation highlights the necessity of proper integration in deriving the electric field from charge distribution.
Necrolunatic
Messages
1
Reaction score
0
Homework Statement
What is electric field at distance $r$ from the center of a sphere of radius $R$ with the charge density $$\rho = \rho_0(R^2 - r^2)$$ where $$\rho_0$$ is a constant and $$r < R$$
Relevant Equations
$$\rho = \rho_0(R^2 - r^2)$$
$$q = \varepsilon_0 \int E.dA$$
My solution is this:
$$q = \varepsilon_0 \int E.dA$$
Based on gauss's law.

Taking the derivative of both sides with respect to $$A$$ we get:
$$\frac{dq}{dA} = \varepsilon_0 E$$
From chain rule:
$$\frac{dq}{dA} = \frac{\frac{dq}{dr}}{\frac{dA}{dr}}$$

On the other hand:
$$q = \int \rho dv = \int \rho \frac{dv}{dr} dr = \int \rho 4 \pi r^2 dr$$
Then the derivative of $$q$$ with respect to $$r$$ is $$\frac{dq}{dr} = \rho 4 \pi r^2$$

And $$A$$ equals $$4 \pi r^2$$, therefor $$\frac{dA}{dr} = 8 \pi r$$
Then we get:
$$\varepsilon_0 E = \frac{dq}{dA} = \frac{\rho 4 \pi r^2}{8 \pi r} = \frac{\rho}{2} r$$

So the electric field at a point within the sphere with distance $$r$$ from the center is
$$E = \frac{\rho}{2 \varepsilon_0} r$$

I don't have the answer to the question but this doesn't seem correct, because in a uniformly charged sphere where $$\rho$$ is $$\frac{Q}{\frac{4}{3} \pi R^3}$$, replacing that into the formula I geta
$$\frac{3 Q}{8 \pi \varepsilon_0 R^3} r$$ which is $$\frac{3}{2}$$ times the actual value $$\frac{Q}{4 \pi \varepsilon_0 R^3} r$$. What am I doing wrong?
 
Physics news on Phys.org
Understand Gauss’s law first. Then integrate instead of taking derivatives.
 
Necrolunatic said:
Taking the derivative of both sides with respect to $$A$$ we get:
$$\frac{dq}{dA} = \varepsilon_0 E$$
This is incorrect. You are effectively differentiating wrt the radius here based on what you do later, but the electric field also depends on the radius so you are missing derivatives.

Much easier is to just note that the rhs is ##\varepsilon_0 EA## and divide both sides by ##\varepsilon_0 A##.
 
  • Like
Likes Necrolunatic
Orodruin said:
Much easier is to just note that the rhs is ##\varepsilon_0 EA## and divide both sides by ##\varepsilon_0 A##.
Wouldn't one have to integrate the lhs to get ##q## before dividing by ##\varepsilon_0 A##?
 
kuruman said:
Wouldn't one have to integrate the lhs to get ##q## before dividing by ##\varepsilon_0 A##?
Obviously you still have to do the integral to get ##q##.
 
  • Like
Likes Necrolunatic
Orodruin said:
Obviously you still have to do the integral to get ##q##.
Judging from OP's initial approach to differentiate both sides, I wanted to make sure that it's also obvious to OP.
 
  • Like
Likes Necrolunatic
Of course, this is integral (🤭) to understanding Gauss’ law.
 
  • Haha
  • Like
Likes Necrolunatic, SammyS, kuruman and 1 other person
Back
Top