SATHEE: Electrostatics Question 752

Electrostatics Question 752

Question: Three identical particles, each possessing the mass m and charge +q, are placed at the corners of an equilateral triangle with side $r_{0} .$ The particles are simultaneously set free and start flying apart symmetrically due to Coulomb’s repulsion foces. The work performed by Coulomb’s forces acting on to a very large distance is $(where k = 1 / 4 π ε_{0} .)$

Options:

A) $\frac{3 k q^{2}}{r_{0}}$

B) $\frac{k q^{2}}{r_{0}}$

C) $\frac{3 k q^{2}}{2 r_{0}}$

D) $\frac{k q^{2}}{2 r_{0}}$

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Answer:

Correct Answer: B

Solution:

[b] Since the given system is closed, the increase in KE is equal to decrease in P.E.

$\Rightarrow \frac{3}{2} m v^{2} = \frac{2 k q^{2}}{r_{0}} - \frac{3 k q^{2}}{r}$

$\Rightarrow v = \sqrt{\frac{2 k q^{2} (r - r_{0})}{m r r_{0}},}$

$\Rightarrow v_{max} = \sqrt{\frac{2 k q^{2}}{m r_{0}}}$

The work performed by the interaction force during the variation of the system’s configuration is equal to the decrease in the potential energy

$W = U_{1} - U_{2} = \frac{3 k q^{2}}{r_{0}}$

$∴$ Work done per particle $= \frac{k q^{2}}{r_{0}}$

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Electrostatics Question 752

Options:

Answer:

Solution:

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Electrostatics Question 752

Options:

Answer:

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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