Two particles A and B, each carrying a charge Q, are held fixed with a separation d between them. A particle C having mass m and charge q is kept at the middle point of the line AB. 1. If it is displaced through a distance x perpendicular to AB, what would be the electric force experienced by it. 2. Assuming x << d, show that this force is proportional to x. 3. Under what conditions will the particle C execute simple harmonic motion if it is released after such a small displacement? Find the time period of the oscillations if these conditions are satisfied.

1. Let Q = charge on A & B Separated by distance d q = charge on c displaced to -AB So, force on 0= F _AB+ F _BO But F_AO =F_BO So, force on ‘0’ in due to vertical component. F =F_AO +F_BO |F_AO|=|F_BO| =2 KQq ( d 2^2+x^2 ) F= 2KQq ( d 2 )^2+x^2 = 4 2 2KQq (d^2+4x^2) x [ ( d 2 )^2+x^2 ]^ 1 2 = 2kQq [ ( d 2 )^2+x^2 ]^ 3 2 x = Electric force 2. When x << d F= 2kQq [ ( d 2 )^2+x^2 ]^ 3 2 x x << d = 2kQq ( d^2 4 )^ 3 2 x a= F m =1 m [ 2kQqx [ ( d^2 4 )+ ^2 ] So time period T=2 g =2 a

Two particles A and B, each carrying a charge Q, are held fixed w…

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Question

Two particles A and B, each carrying a charge Q, are held fixed with a separation d between them. A particle C having mass m and charge q is kept at the middle point of the line AB.

If it is displaced through a distance x perpendicular to AB, what would be the electric force experienced by it.
Assuming x << d, show that this force is proportional to x.
Under what conditions will the particle C execute simple harmonic motion if it is released after such a small displacement?

Find the time period of the oscillations if these conditions are satisfied.

✓

Answer

Let Q = charge on A & B Separated by distance d

q = charge on c displaced $\bot$ to -AB

So, force on $0=\overline{\text{F}}_\text{AB}+\overline{\text{F}}_\text{BO}$

But $\text{F}_\text{AO}\cos\theta=\text{F}_\text{BO}\cos\theta$

So, force on ‘0’ in due to vertical component.

$\overline{\text{F}}=\text{F}_\text{AO}\sin\theta+\text{F}_\text{BO}\sin\theta$ $|\text{F}_\text{AO}|=|\text{F}_\text{BO}|$

$=2\frac{\text{KQq}}{\Big(\frac{\text{d}}{2^2+\text{x}^2}\big)}\sin\theta$

$\text{F}=\frac{2\text{KQq}}{\Big(\frac{\text{d}}{2}\Big)^2+\text{x}^2}\sin\theta$

$=\frac{4\times2\times2\text{KQq}}{(\text{d}^2+4\text{x}^2)}\times\frac{\text{x}}{\Big[\big(\frac{\text{d}}{2}\big)^2+\text{x}^2\Big]^{\frac{1}{2}}}$

$=\frac{2\text{kQq}}{\Big[\big(\frac{\text{d}}{2}\big)^2+\text{x}^2\Big]^{\frac{3}{2}}}\text{x}$ = Electric force $\Rightarrow\text{F}\propto\text{x}$

When x << d

$\text{F}=\frac{2\text{kQq}}{\Big[\big(\frac{\text{d}}{2}\big)^2+\text{x}^2\Big]^{\frac{3}{2}}}\text{x}$ x << d

$\Rightarrow\text{F}=\frac{2\text{kQq}}{\Big(\frac{\text{d}^2}{4}\Big)^{\frac{3}{2}}}\text{x}\Rightarrow\text{F}\propto\text{x}$

$\text{a}=\frac{\text{F}}{\text{m}}=\frac{1}{\text{m}}\Bigg[\frac{2\text{kQqx}}{\Big[\big(\frac{\text{d}^2}{4}\big)+\ell^2}\Bigg]$

So time period $\text{T}=2\pi\sqrt{\frac{\ell}{\text{g}}}$

$=2\pi\sqrt{\frac{\ell}{\text{a}}}$

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