SHM due to Charges

Various SHM and No SHM possibilities due to Charges

SHM in the line joining two positive charges

Positive charge in the middle displaced by a small distance x along axial line.

\[F_1=\frac{kQq}{(l-x)^2}\] \[F_2=\frac{kQq}{(l+x)^2}\] \[F_{\text{restoring}}=F_1-F_2\] \[F_{\text{restoring}}=kQq\left[\frac{1}{(l-x)^2}-\frac{1}{(l+x)^2}\right]\] \[ma=-\,kQq\,\frac{4Lx}{\left(l^2-x^2\right)^2}\] For small displacement \(x \ll l:\) \[a=-\frac{4kQq}{ml^3}\,x\] \[a=-\omega^2x\] \[\omega=\sqrt{\frac{4kQq}{m\,l^3}}\] \[T = \frac{2\pi}{\omega}= 2\pi \sqrt{\frac{ml^3}{4kQq}}\]

SHM in the Equatorial line between two positive charges

Negative charge in the middle displaced by a small distance x along Equatorial line.

\[F_{\text{restoring}} = 2F \sin\theta\] \[F = 2 \cdot \frac{kQq}{(x^2 + l^2)} \cdot \frac{x}{\sqrt{x^2 + l^2}}\] \[ma = - \frac{2kQq}{(x^2 + l^2)^{3/2}} x\] For small displacement \(x \ll l:\) \[a = - \frac{2kQq}{ml^3} x\] \[a = -\omega^2 x\] \[\omega = \sqrt{\frac{2kQq}{ml^3}}\] \[T = \frac{2\pi}{\omega}= 2\pi \sqrt{\frac{ml^3}{2kQq}}\]

No SHM when a positive charge is diplaced in the Equatorial line

Location of Maximum Force using Differentiation

\[F_{\text{net}}=2F\sin\theta\] \[F_{\text{net}}=2\left(\frac{kQq}{x^2+l^2}\right)\left(\frac{x}{\sqrt{x^2+l^2}}\right)=\frac{2kQq\,x}{(x^2+l^2)^{3/2}}\] \[\text{To find }F_{\max},\qquad\frac{dF}{dx}=0\] \[2kQq\left[\frac{(x^2+l^2)^{3/2}(1)-x\left(\frac{3}{2}\right)(x^2+l^2)^{1/2}(2x)}{\left[(x^2+l^2)^{3/2}\right]^2}\right]=0\] \[(x^2+l^2)-3x^2=0\] \[l^2-2x^2=0\] \[x=\frac{l}{\sqrt{2}}\]

SHM of a Dipole in an External Electric Field

\[\tau_{\text{restoring}} = pE\sin\theta\] \[I\alpha = -pE\,\theta\] \[\alpha = -\frac{pE}{I}\theta\] \[\alpha = -\omega^2\theta\] \[\omega = \sqrt{\frac{pE}{I}}\] \[T=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{I}{pE}}\]

No SHM when a positive charge is diplaced in the Axial line of the Charged Ring Q

Location of Maximum Force using Differentiation

\[E_{\text{axial}}=\frac{kQ\,x}{(x^2+R^2)^{3/2}}\] \[F_{\text{net}}=\frac{kQq\,x}{(x^2+R^2)^{3/2}}\] \[\text{To find }F_{\max},\qquad\frac{dF}{dx}=0\] \[kQq\left[\frac{(x^2+R^2)^{3/2}(1)-x\left(\frac{3}{2}\right)(x^2+R^2)^{1/2}(2x)}{\left[(x^2+R^2)^{3/2}\right]^2}\right]=0\] \[(x^2+R^2)-3x^2=0\] \[R^2-2x^2=0\] \[x=\frac{R}{\sqrt{2}}\]

SHM when a negative charge is diplaced in the Axial line of the Charged Ring Q

Negative charge q in the center displaced by a small distance x along axial line.

\[E_{\text{axial}}=\frac{kQ\,x}{(x^2+R^2)^{3/2}}\] \[F_{\text{restoring}}=\frac{kQq\,x}{(x^2+R^2)^{3/2}}\] \[ma = - \frac{kQq}{(x^2 + R^2)^{3/2}} x\] For small displacement \(x \ll R:\) \[a = - \frac{kQq}{mR^3} x\] \[a = -\omega^2 x\] \[\omega = \sqrt{\frac{kQq}{mR^3}}\] \[T = \frac{2\pi}{\omega}= 2\pi \sqrt{\frac{mR^3}{kQq}}\]

No SHM when a negative charge is diplaced in the Axial line of the Charged Disk Q

Negative charge q in the center displaced by a small distance x along axial line.

\[E = k\,\frac{2Qx}{R^2}\left(\frac{1}{x}-\frac{1}{\sqrt{R^2+x^2}}\right)\] \[F_{\text{restoring}}=k\,\frac{2Qqx}{R^2}\left(\frac{1}{x}-\frac{1}{\sqrt{R^2+x^2}}\right)\] \[ma = - k\,\frac{2Qqx}{R^2}\left(\frac{1}{x}-\frac{1}{\sqrt{R^2+x^2}}\right)\] For small displacement \(x \ll R:\) \[a = - k\,\frac{2Qqx}{mR^2}\left(\frac{1}{x}-\frac{1}{R}\right) \] \[a = - \,\frac{k2Qq}{mR^2} + \frac{k2Qq}{mR^3} x \] We see that the coefficient of x in the above equation obtained is having positive sign, so it is not a SHM. Also Compare it with \[y=mx+c\] we see that the slope is positive. Which means it does not follow the characteristic equation of SHM. \[a = -\omega^2x\] \[a = -\omega^2\left(x-x_o\right)\] for shifted origin.

SHM when a negative charge is diplaced in the Axial line of the Charged Disk Q with an Annular Region

Negative charge q in the center displaced by a small distance x along axial line.

In the small shaded region, \[\sigma=\frac{dq}{2\pi r \,dr}\] From the E due to a charged Ring, \[dE=\frac{k\,dq\,x}{\left(r^2+x^2\right)^{3/2}}\] \[E=k\int_{r_1}^{r_2}\frac{\sigma\,2\pi r\,dr\,x}{\left(r^2+x^2\right)^{3/2}}\] \[E=k\,\sigma\,2\pi x\int_{r_1}^{r_2}\frac{r\,dr}{\left(r^2+x^2\right)^{3/2}}\] \[E=k\,\sigma\,2\pi x\left[-\frac{1}{\sqrt{r^2+x^2}}\right]_{r_1}^{r_2}\] \[E=k\,\sigma\,2\pi x\left(\frac{1}{\sqrt{r_1^2+x^2}}-\frac{1}{\sqrt{r_2^2+x^2}}\right)\] For small displacement \(x \ll r:\) \[E=k\,\sigma\,2\pi x\left(\frac{1}{r_1}-\frac{1}{r_2}\right)\] Force on a charge $q_o$ \[F=k\,\sigma\,2\pi q_0\left(\frac{1}{r_1}-\frac{1}{r_2}\right)x\] \[ma=-k\,\sigma\,2\pi q_0\left(\frac{1}{r_1}-\frac{1}{r_2}\right)x\] \[a=-\frac{k\,\sigma\,2\pi q_0}{m}\left(\frac{1}{r_1}-\frac{1}{r_2}\right)x\] \[a=-\omega^2x\] \[\omega=\sqrt{\frac{k\sigma\,2\pi q_0}{m}\left(\frac{1}{r_1}-\frac{1}{r_2}\right)}\] \[T=\frac{2\pi}{\omega}=\frac{2\pi}{\sqrt{\frac{k\sigma\,2\pi q_0}{m}\left(\frac{1}{r_1}-\frac{1}{r_2}\right)}}\]

Geogebra Link for Graphs and Edits https://www.geogebra.org/graphing/guywhhke

Questions from JEE NEET Past Year Papers

For a particle of mass 1 mg and charge q is lying at the mid-point of two stationary particles kept at a distance 2 m when each is carrying same charge q. If the free charged particle is displaced from its equilibrium position through distance 'x' ( x  very less than 1 m). The particle executes SHM. Its angular frequency of oscillation will be ______ x 10^5 rad/sec if q^2 = 10 C^2.

[ 6000]

Two identical positive charge Q each are fixed at a distance of 2a apart from each other. Another point charge q_o with mass m is placed at midpoint between two fixed charges. For a small displacement along the line joining the fixed charges, the charge q_o executes SHM. The time period of oscillation of charge q_o will be ..........

[Ans: sqrt(4 pi ^2 epsilon_o m a^3 / q_o Q ) ]