Question

Repeat the previous problem if the particle C is displaced through a distance x along the line AB.

Answer 1

Explanation:

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Answer 2

Explanation:

Step 1:

It is given that

$F_{A C}=\frac{K Q q}{(\ell+x)^{2}} \quad F_{C A}=\frac{K Q q}{(\ell-x)^{2}}$

So, the net force

$=\frac{1}{4 \pi \epsilon_{0}}\left[\frac{Q q}{\left(\frac{d}{2}-x\right)^{2}}-\frac{Q q}{\left(\frac{d}{2}+x\right)^{2}}\right]$

$=\frac{Q q}{4 \pi \epsilon_{0}} \frac{\left[\left(\frac{d}{2}\right)^{2}+x^{2}+x d-\left(\frac{d}{2}\right)^{2}-x^{2}+x d\right]}{\left[\left(\frac{d}{2}\right)^{2}-x^{2}\right]^{2}}$

Step 2:

when, x << d

So net force    = $\frac{q Q}{4 \pi \in_{0}} \frac{(2 x d)}{d^{4}}$

                       $=\frac{q Q}{4 \pi \epsilon_{0}} \frac{2 x}{d^{3}}$

Or   $m\left(\frac{2 \pi}{\mathrm{T}}\right)^{2} x=\frac{2 x q Q}{4 \pi \epsilon_{0} d^{3}}$

     $\text { Net force }=\mathrm{KQq}\left[\frac{1}{(\ell-\mathrm{x})^{2}}-\frac{1}{(\ell+\mathrm{x})^{2}}\right]$

                     $=\mathrm{KQq}\left[\frac{(\ell+\mathrm{x})^{2}-(\ell-\mathrm{x})^{2}}{(\ell+\mathrm{x})^{2}(\ell-\mathrm{x})^{2}}\right]=\mathrm{KQq}\left[\frac{4 \ell \mathrm{x}}{\left(\ell^{2}-\mathrm{x}^{2}\right)^{2}}\right]$

$\text { net } F=\frac{K Q q 4 \ell x}{\ell^{4}}=\frac{K Q q 4 x}{\ell^{3}} \quad \text { acceleration }=\frac{4 \mathrm{KQqx}}{\mathrm{m} \ell^{3}}$

Step 3:

Time period = $Time period =2 \pi \sqrt{\frac{\text { displacement }}{\text { acceleration }}}$

= $2 \pi \sqrt{\frac{\mathrm{xm} \ell^{3}}{4 \mathrm{KQqx}}}=2 \pi \sqrt{\frac{\mathrm{m} \ell^{3}}{4 \mathrm{KQq}}}$

= $\sqrt{\frac{4 \pi^{2} m \ell^{3} 4 \pi \varepsilon_{0}}{4 Q q}}=\sqrt{\frac{4 \pi^{3} m \ell^{3} \varepsilon_{0}}{Q q}}$          

                       $=\sqrt{4 \pi^{3} \mathrm{md}^{3} \varepsilon_{0} 8 Q q}$  

                        =  $\left[\frac{\pi^{3} \mathrm{md}^{3} \varepsilon_{0}}{2 \mathrm{Qq}}\right]^{1 / 2}$  

It is to be noted that the distance is an arithmetical dimension of how far apart points or objects are. In everyday usage or physics, the distance is referred as an estimation or a physical length on the basis of other criteria.