Testing Coulomb's law

Because the first part $\frac{kq_1q_2}{r^2}$ of the force does no force to any charge inside the shell, we only need to consider the second part $\delta F(r)$ of the force.

Suppose when the distance from the charge q to O is x (x is very small).

Suppose the shell is made up of many very small ring, whose centers are in the same diameter.

Consider one ring of them, which is limited from z to z+dz, has radius $r=\sqrt{R^2-z^2}$ , distance to charge q is $l=\sqrt{r^2+(z-x)^2} \approx \sqrt{R^2-2zx}$ and charge $dQ=\frac{Qdz}{2R}$ .

The force by this ring to charge q: $dF=\frac{k'qdQ}{l^3} \frac{z-x}{l}=\frac{k'qQdz(z-x)}{2Rl^4}$

$=\frac{k'qQdz(z-x)}{2R(R^2-2zx)^2} \approx \frac{k'qQdz(z-x)}{2R^5} (1+\frac{4zx}{R^2})$

$\approx \frac{k'qQdz}{2R^5} (z-x+\frac{4z^2x}{R^2})$

Therefore, the total force into the charge q is: $F=\int \limits_{-R}^R \frac{k'qQdz}{2R^5} (z-x+\frac{4z^2x}{R^2})$

$=\frac{k'qQx}{3R^4}$ .

Therefore, the period: $T=2\pi \sqrt{\frac{3mR^4}{k'qQ}}=\sqrt{3} T_0$ , so $c=\sqrt{3}$ .

The electrostatic potential for a single point charge $q$ must be $\frac{kq}{r} + \frac{k'q}{2r^2}$ if we are to get the right force equation. Thus the potential at a point $P$ inside the sphere, where $OP=r<R$ , is (writing $r_1^2 = R^2+r^2-2Rr\cos\theta$ ): $\begin{array}{rcl} \phi(r) & = & \int_{-\pi}^\pi \frac{2\pi R^2\sin\theta\,d\theta}{4\pi R^2} Q \Big(\frac{k}{r_1} + \frac{k'}{2r_1^2}\Big) \\ & = & \tfrac{Q}{2}\int_{-1}^1 \Big(\frac{k}{\sqrt{R^2+r^2-2Rru}} + \frac{k'}{2(R^2+r^2-2Rru)}\Big)\,du \\ & = & \frac{kQ}{2R} + \frac{k'Q}{4Rr}\ln\Big(\frac{R+r}{R-r}\Big) \\ & = & \frac{kQ}{2R} + \frac{k'Q}{2R}\sum_{n=0}^\infty \frac{r^{2n}}{(2n+1)R^{2n+1}} \\ & \approx & \frac{kQ}{2R} + \frac{k'Q}{2R^2} + \frac{k'Q}{6R^4}r^2 + \cdots \end{array}$ and hence the central force on a particle of charge at the point $P$ is $-q\frac{d\phi}{dr} \; \approx \; -\frac{k'Qq}{3R^4}r \; = \; -\frac{|\delta F(R)|}{3R}r$ Thus a particle of charge $q$ obeys the following approximate differential equation for small oscillations about $O$ : $\ddot{r} + \frac{|\delta F(R)|}{3mR}r \; = \; 0$ and so the particle performs approximate SHM of period $2\pi \sqrt{\frac{3mR}{|\delta F(R)|}} \; = \; T_0\sqrt{3}$

Clearly , the $F_{net}$ due to traditional coulomb's law would still be zero. Now , we have to check $F_{net}$ due to deviation .

First consider a ring having charge $q$ of radius $r$

Deviation for field due to the ring at a distance $y$ = $\frac{k^{'}qy}{(r^2 + y^2)^2}$ ....... $(i)$ (its easy , prove yourself).

Now, consider this example.

Let us consider a ring on the sphere subtending an angle $\theta$ at the center.

$y_{for ring} = Rcos\theta + x$ , where $x$ is the amount by which $q$ has been displaced, and radius $r = Rsin\theta$ .

$dq_{ring} = \frac{Q}{4 \pi R^2}2\pi R sin\theta Rd\theta = \frac{Q sin\theta d\theta}{2}$ .

Now , $dF = \frac{k^{'}(Q sin\theta d\theta)q(R cos\theta + x)}{2( (R + xcos\theta )^2 + (Rsin\theta)^2)^2}$ [Using $(i)$ ]

= $\frac{k^{'}(Q sin\theta d\theta)q(R cos\theta + x)}{2( R^2 + x^2 + 2Rxcos \theta)^2}$ .

Using $(1 + x )^n = 1 + nx$ for $x << 1$ for the denominator , We get :

$dF = \frac{k^{'}qQ}{R^4} (Rcos \theta + x)( 1 - 4\frac{x}{R} cos\theta)sin\theta d\theta$

$\Rightarrow F = \int_{0}^{\pi} \frac{k^{'}qQ}{R^4} (Rcos \theta + x)( 1 - 4\frac{x}{R} cos\theta)sin\theta d\theta$

Put $\cos \theta = t$ to get :

$F = \int_{-1}^{1} \frac{k^{'}qQ}{R^4} (Rt + x)( 1 - 4\frac{x}{R}t)dt$

$\Rightarrow F = -\frac{k^{'}qQ}{3 R^4} = -\frac{\delta(F(R)}{3R}$

$\Rightarrow T = \sqrt{\frac{3mR}{|\delta(F(R)|}} = \sqrt3 T_{0}$

Hence , $c = \sqrt{3} = \fbox{1.732}$