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 speed (in SI units) of revolving particle when it reaches point A, will be? Use $(K = \frac{1}{4 π ε _{0}})$ and $(\frac{KQq}{Ma} = 1000 S . I$ units $)$

Answer:

10.00

Solution:

Total energy of the orbit: $E = - \frac{K Qq}{5 a}$ . At A (distance 4a):

$\frac{1}{2} M V_{A}^{2} - \frac{K Qq}{4 a} = - \frac{K Qq}{5 a} \Rightarrow \frac{1}{2} M V_{A}^{2} = \frac{K Qq}{20 a}$ .

$V_{A}^{2} = \frac{1}{10} \frac{K Qq}{M a} = \frac{1000}{10} = 100 \Rightarrow V_{A} = 10 m / s$ .

Stream:JEESubject:PhysicsTopic:ElectrostaticsSubtopic:Miscellaneous

⏱ 3mℹ️ Source: QB

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