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

• Necrolunatic
Necrolunatic
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?

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##.

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##.

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.

Necrolunatic
Of course, this is integral () to understanding Gauss’ law.

Necrolunatic, SammyS, kuruman and 1 other person

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