- #1
Necrolunatic
- 1
- 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?
$$q = \varepsilon_0 \int E.dA$$
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?
) to understanding Gauss’ law.