# Electric field due to a solid sphere of charge

I have been trying to understand the last step of this derivation. Consider a sphere made up of charge $+q$. Let $R$ be the radius of the sphere and $O$, its center.

A point $P$ lies inside the sphere of charge. In such a case, Gaussian surface is a spherical shell,whose center is $O$ and radius is $r$ (=OP). If $q'$ is charge enclosed by Gaussian surface,then

$$E\times4\pi r^2=q'/\epsilon$$ where $\epsilon=$ absolute permittivity of free space.

$$E=\frac{1}{4\pi\epsilon}\times(q'/r^2)$$ for $(r<R)$

$$q'=\frac{q}{\frac{4}{3}\pi R^3}\times\frac{4}{3}\pi r^3=\frac{qr^3}{R^3}$$

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Hello Drown, Welcome to Physics.SE. Here, we make use of TeX - an useful resource for formulas. Please have a look my revision of your question. It would be useful if you could modify the content to ask specifically what you require from now... –  Waffle's Crazy Peanut Oct 25 '12 at 12:44

I presume your problem is the calculation of $q'=\frac{r^3}{R^3}q$.
This is perhaps easier to explain by splitting the calculation in two steps. The solid ball of charge is supposed to be homogeneous, so it has a charge density $$\rho=\frac{\textrm{total charge}}{\textrm{total volume}}=\frac{q}{\frac{4\pi}{3}R^3}.$$ The smaller sphere has volume $V_r=\frac{4\pi}{3}r^3$, and therefore has charge $$q'=\rho V_r=\frac{q}{\frac{4\pi}{3}R^3}\frac{4\pi}{3}r^3=\frac{r^3}{R^3}q.$$
That $\rho$ is called charge density which is the charge per unit volume. $\rho=q/V$. This is the same as linear and surface charge densities... –  Waffle's Crazy Peanut Oct 25 '12 at 12:58