Name: Surface and Volume Charge Distribution
Uploaded: 2019-07-19
Duration: 4 min 37 s
Description: This video helps students to understand surface and volume charge distribution. Here, we will also define surface and volume charge densities

Question

Two charges $- q$ each are fixed separated by distance $2 d$ . A third charge $q$ of mass $m$ placed at the mid-point is displaced slightly by $x (x < < d)$ perpendicular to the line joining the two fixed charged as shown in Figure. Show that $q$ will perform simple harmonic oscillation of time period.

$T = {[\frac{8 π^{3} ε_{0} {md}^{3}}{q^{2}}]}^{1 / 2}$

Question Image

$T = {[\frac{8 π^{3} ε_{0} {md}^{3}}{q^{2}}]}^{1 / 2}$

Accepted Answer

Force acting on charge $q$ due to $2 d$ at A will be $q$ along $P$ to $A$

So, $F_{R,} = \frac{1}{4 {πε}_{0}} \frac{q_{1} q_{2}}{r^{2}} = \frac{1}{4 {πε}_{0}} \cdot \frac{q (- q)}{r^{2}}$

$F_{PA} = \frac{- q^{2}}{4 {πε}_{0} r^{2}}$

Similarly,

$F_{PB} = \frac{- q^{2}}{4 {πε}_{0} r^{2}} [∵ |F_{R_{A}}| = |F_{PB}| = F]$

So, the horizontal components of $F_{PA}$ and $F_{PB}$ are equal and opposite so neutralise each other. The vertical components of $F_{PA}$ and $F_{PB}$ are downward and so add up.

Net force on $q$ at $P$ is $F_{PB} cosθ + F_{PA} cosθ = 2 Fcosθ [∵ |F_{PA}| = F_{PB} | = F]$

$= \frac{- 2 \cdot q^{2}}{4 {πε}_{0} r^{2}} cosθ$

Force on $q$ downwards $= \frac{- 2 . q^{2}}{4 {πε}_{0} r^{2}} \frac{x}{r}$

$r^{2} = x^{2} + d^{2}$ [by Pythagoras theorem]

Force on $q = = \frac{- 2 . q^{2}}{4 {πε}_{0}} \frac{x}{{(x^{2} + d^{2})}^{3 / 2}}$

Then, $x < < d \Rightarrow x^{2} < < d^{2}$

As negative sign shows force of attraction.

So net force on $q$ at $P$ downwards $= (\frac{- 2 q^{2}}{4 {πε}_{0} d^{3}}) x$

So the force on $q$ is directly proportional to the displacement from mean position $O$ (the midpoint of the segment joining $- q$ and - $q$ charges). So the motion of $q$ about $O$ will be S.H.M.

$F = - kx$ where $k = \frac{2 q^{2}}{4 {πε}_{0} d^{3}}$

So, $ω = \sqrt{\frac{k}{m}}$

$T = \frac{2 π}{ω} = 2 π \sqrt{\frac{m}{k}}$

$T = 2 π \sqrt{\frac{4 {πmε}_{0} d^{3}}{2 q^{2}}}$

$T = {[\frac{8 π^{3} ε_{0} {md}^{3}}{q^{2}}]}^{\frac{1}{2}}$

Hence, proved.

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