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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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    $\begingroup$ 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... $\endgroup$ – Waffle's Crazy Peanut Oct 25 '12 at 12:44
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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 $$q'=\rho V_r=\frac{q}{\frac{4\pi}{3}R^3}\frac{4\pi}{3}r^3=\frac{r^3}{R^3}q.$$

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  • $\begingroup$ From where did you get this formula q′=ρVr. While studying i never got this formula. $\endgroup$ – Drownpc Oct 25 '12 at 12:51
  • $\begingroup$ 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... $\endgroup$ – Waffle's Crazy Peanut Oct 25 '12 at 12:58
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enter image description hereElectric field due to solid sphere.This is so easy.

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  • $\begingroup$ Hi and welcome to Physics SE! Please do not post screenshots as answers. Write equations using MathJax so that search engines can index this post. $\endgroup$ – exp ikx May 17 at 12:12

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