# Charge and Electric field inside a sphere

Suppose we are given a sphere with total charge $$Q$$ and radius $$R$$. To find the total charge $$Q$$, we can use the relationship $$\rho = \frac{dQ}{dV}\implies Q=\int_0^R\rho \cdot dVdr$$

where $$dV$$ is the derivative of volume with respect to $$r$$, the distance from the center of the sphere, and $$\rho$$ is the volume charge density. Now suppose we want to find the electric field $$E$$ inside the sphere. From the definition of flux, we obtain $$\Phi=EA=\frac{Q}{\epsilon_0}$$

But in this instance, the bounds of integration are different, we have that $$EA=\frac{1}{\epsilon_0}\int_0^r\rho\cdot dV dr$$

I don't see why the relationship $$Q=\int \rho dV dr$$ differs between these two instances - I suspect it might have something to do with the flux and maybe in the latter case we are not dealing with the total charge in the sphere, but instead the amount of charge passing through some infinitesimal surface within the sphere - given by $$r$$ instead of the radius $$R$$. Can anyone clear this up for me?

The charge density $$\rho(x,y,z)$$ is the charge per unit volume. So when you want to get the total charge of a given density in a Volume $$V$$ you have to integrate the density over the volume.

$$$$Q = \int_V \rho(x,y,z) dxdydz = \int_0^{\pi} d\theta \int_0^{2\pi} d\phi \int_0^{R} dr \rho(r,\phi,\theta) r^2 \sin(\theta)$$$$

You added the $$dr$$ which is already in the $$dV = r^2 \sin(\theta) dr d\phi d\theta$$.

And for the latter part, I can't see where you get the different integration limit from. You can look up Poisson Equation and Gauss's law. From that you see

$$$$\frac{Q}{\epsilon_0} = \frac{1}{\epsilon_0} \int_V \rho dV = \int_V \vec{\nabla} \vec{E} dV = \int_A \vec{E} d\vec{A}$$$$

Where Gauss's law is used in the last step and Poisson Equation in the second to last. Hope that somehow helps and was not too basic. Cheers

First of all, from the formula you have written for Q, it's dimensions are not of charge, but it's dimensions is charge*length

You have written extra dr in the integration formula.

$$\rho$$ is volume charge density and to make it charge, you just have multiply it by volume. (Note- if you want to denote, that volume $$V$$ as a function of $$r$$, then you can use $$d^3 (r)$$)

Now considering your question, the R, is used for radius of sphere, while r is variable, because if you select your Gaussian surface inside the sphere, then the total charge enclosed by it will be somewhat less than total charge enclosed by it.