Gujarati
Hindi
1. Electric Charges and Fields
normal

A charged particle with charge $q$ and mass $m$ starts with an initial kinetic energy $K$ at the centre of a uniformly charged spherical region of total charge $Q$ and radius $R$. Charges $q$ and $Q$ have opposite signs. The spherically charged region is not free to move and kinetic energy $K$ is just sufficient for the charge particle to reach boundary of the spherical charge. How much time does it take the particle to reach the boundary of the region?

A

$\sqrt[\pi ]{{\frac{{4\pi {\varepsilon _o}m{R^3}}}{{qQ}}}}$

B

$\sqrt[{\frac{\pi }{2}}]{{\frac{{4\pi {\varepsilon _o}m{R^3}}}{{qQ}}}}$

C

$\sqrt[{\frac{\pi }{4}}]{{\frac{{4\pi {\varepsilon _o}m{R^3}}}{{qQ}}}}$

D

None of the above.

Solution

$F=-\frac{\rho q}{3 \epsilon_{0}} r$

Therefore it excutes $SHM.$ If remains always within the sphere.

$\frac{{{{\rm{d}}^2}{\rm{r}}}}{{{\rm{d}}{{\rm{t}}^2}}} =  – \frac{{\rho {\rm{q}}}}{{3{ \in _0}{\rm{m}}}}{\rm{r}}$

$\Rightarrow$ Angular frequency

$\omega  = \sqrt {\frac{{\rho q}}{{3{ \in _0}m}}} $

Time to reach from centre to surface $\mathrm{t}=\frac{\mathrm{T}}{4}$

${\rm{t}} = \frac{{\rm{T}}}{4} = \frac{1}{4}\frac{{2\pi }}{\omega } = \frac{\pi }{{2\omega }} = \frac{\pi }{2}\sqrt {\frac{{3{ \in _0}{\rm{m}}}}{{\rho {\rm{q}}}}} $         ………$(i)$

We know that $\rho=\frac{3 \mathrm{Q}}{4 \pi \mathrm{R}^{3}}$

so from $(i)$ $t = \frac{\pi }{2}\sqrt {\frac{{4\pi { \in _0}{\rm{m}}{{\rm{R}}^3}}}{{{\rm{Qq}}}}} $

Standard 12
Physics

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