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1. Electric Charges and Fields
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$M$ દળનો વિદ્યુતભાર $q$ એ $q$ વિદ્યુતભારની આજુબાજુ સ્થિત વિદ્યુત આકર્ષણને લીધે પરિભ્રમણ કરે છે. તેની ગતિનો આવર્તકાળ..... સૂત્રની મદદથી આપી શકાય છે.
A${T^2}\,\, = \,\,\frac{{11{\pi ^3}\,\,{ \in _0}\,\,m{R^2}}}{{Qq}}$
B${T^2}\,\, = \,\,\frac{{16{\pi ^3}\,\,{ \in _0}\,\,m{R^3}}}{{Qq}}$
C${T^2}\,\, = \,\,\frac{{16{\pi ^4}\,\,{ \in _0}\,\,m{R^2}}}{{Qq}}$
D${T^2}\,\, = \,\,\frac{{18{\pi ^3}\,\,{ \in _0}\,\,m{R^4}}}{{Qq}}$
Solution
$Q$ ની વર્તુળીય ગતિ માટે
${F_{CP}}\,\, = \,\,{F_e}\,\, \Rightarrow \,\,\,\,\,\frac{{m{v^2}}}{R}\,\,\, = \,\,\frac{{kQq}}{{{R^2}}}\,\, \Rightarrow \,\,v\,\, = \,\sqrt {\frac{{kQq}}{{mR}}} $
$⇒$આવર્તકાળ ${\text{T}}\,\, = \,{\text{ }}\frac{{{\text{2}}\pi {\text{R}}}}{{\text{v}}}\,\,\, = \,\,\frac{{2\pi R\sqrt {mR} }}{{\sqrt {kQq} }}\,\,\, \Rightarrow \,\,{T^2}\, = \,\,\,\frac{{4{\pi ^2}{R^2}(mR)4\pi \,\,{ \in _0}}}{{(Qq)}}\,\,\, = \,\,\frac{{16{\pi ^3}\,\,{ \in _0}\,\,m{R^3}}}{{Qq}}$
${F_{CP}}\,\, = \,\,{F_e}\,\, \Rightarrow \,\,\,\,\,\frac{{m{v^2}}}{R}\,\,\, = \,\,\frac{{kQq}}{{{R^2}}}\,\, \Rightarrow \,\,v\,\, = \,\sqrt {\frac{{kQq}}{{mR}}} $
$⇒$આવર્તકાળ ${\text{T}}\,\, = \,{\text{ }}\frac{{{\text{2}}\pi {\text{R}}}}{{\text{v}}}\,\,\, = \,\,\frac{{2\pi R\sqrt {mR} }}{{\sqrt {kQq} }}\,\,\, \Rightarrow \,\,{T^2}\, = \,\,\,\frac{{4{\pi ^2}{R^2}(mR)4\pi \,\,{ \in _0}}}{{(Qq)}}\,\,\, = \,\,\frac{{16{\pi ^3}\,\,{ \in _0}\,\,m{R^3}}}{{Qq}}$
Standard 12
Physics
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