SATHEE: Electrostatics Question 250

Electrostatics Question 250

Question: An elementary particle of mass $m$ and charge $+ e$ is projected with velocity $v$ at a much more massive particle of charge $Z e,$ where $Z > 0.$ What is the closest possible approach of the incident particle [Orissa JEE 2002]

Options:

A) $r^{2}$

B) r

C) $\frac{1}{r}$

D) $\frac{1}{r^{2}}$

Show Answer

Answer:

Correct Answer: B

Solution:

$E_{x} = - \frac{d V}{d x} = - k y$ ;

$E_{y} = - \frac{d V}{d y} = - k x$

therefore $E = \sqrt{E_{x}^{2} + E_{y}^{2}} = k \sqrt{x^{2} + y^{2}} = k r$

therefore E µ r

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अगला

Electrostatics Question 250

Question: An elementary particle of mass $m$ and charge $+ e$ is projected with velocity $v$ at a much more massive particle of charge $Z e,$ where $Z > 0.$ What is the closest possible approach of the incident particle [Orissa JEE 2002]

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

Electrostatics Question 250

Question: An elementary particle of mass m and charge +e is projected with velocity v at a much more massive particle of charge Ze, where Z>0. What is the closest possible approach of the incident particle [Orissa JEE 2002]

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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Question: An elementary particle of mass $m$ and charge $+ e$ is projected with velocity $v$ at a much more massive particle of charge $Z e,$ where $Z > 0.$ What is the closest possible approach of the incident particle [Orissa JEE 2002]