Physics - Electrostatics Question with Solution | TestHub

PhysicsElectrostaticsMiscellaneousMedium3 minQB

PhysicsMediumnumerical

Passage / Comprehension

Consider a situation in which a negatively charged particle of mass M and charge -q revolves in elliptical path around a fixed charge Q. The closest and farthest distance of the moving particle from fixed charge is a and 4 a respectively. Point A and B are the two extreme points of the major axis of ellipse as shown in the figure. Charge Q is present at one of the Foci of the ellipse.\\

The radius of curvature of path (in SI units) of revolving particle at point B , will be :- Use [ $a = 25$ (S.I units)]

Answer:

40.00

Solution:

At B (distance a): $\frac{1}{2} M V_{B}^{2} - \frac{K Qq}{a} = - \frac{K Qq}{5 a} \Rightarrow V_{B}^{2} = \frac{8}{5} \times 1000 = 1600$ .

$\frac{M V _{B}^{2}}{ρ _{B}} = \frac{K Qq}{a ^{2}} \Rightarrow ρ_{B} = \frac{M V _{B}^{2} a ^{2}}{K Qq} = \frac{1600 \times 25}{1000} = 40 m$ .

Stream:JEESubject:PhysicsTopic:ElectrostaticsSubtopic:Miscellaneous

⏱ 3mℹ️ Source: QB

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