4
$\begingroup$

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 $(r<R)$

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

$\endgroup$
1
  • 1
    $\begingroup$ 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... $\endgroup$ Oct 25, 2012 at 12:44

3 Answers 3

1
$\begingroup$

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 $$q'=\rho V_r=\frac{q}{\frac{4\pi}{3}R^3}\frac{4\pi}{3}r^3=\frac{r^3}{R^3}q.$$

$\endgroup$
2
  • $\begingroup$ From where did you get this formula q′=ρVr. While studying i never got this formula. $\endgroup$
    – Drownpc
    Oct 25, 2012 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... $\endgroup$ Oct 25, 2012 at 12:58
0
$\begingroup$

$$E= \frac{kQr}{R^3} \\ = \frac{\rho r}{3 \epsilon_0}$$

$\endgroup$
1
  • $\begingroup$ Hi, it's best to use MathJax to format your posts for both clarity and readability. When writing answers, it is also best to give an explanation rather than just an answer. $\endgroup$ Jul 14, 2021 at 3:18
-3
$\begingroup$

enter image description hereElectric field due to solid sphere.This is so easy.

$\endgroup$
2
  • $\begingroup$ 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, 2019 at 12:12
  • 1
    $\begingroup$ If its so easy how did u get the completely wrong answer LOL $\endgroup$ Mar 28, 2021 at 16:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.