SATHEE: Electrostatics Question 736

Electrostatics Question 736

Question: A system consists of a uniform charged sphere of radius R and a surrounding medium filled by a charge with the volume density $ρ = \frac{α}{r},$ where $α$ a positive constant is and r is the distance from the center of the charge. The charge of the sphere for which the electric field intensity E outside the sphere is independent of r is-

Options:

A) $π R^{2} α$

B) $4 π R^{2} α$

C) $2 π R^{2} α$

D) $3 π R^{2} α / 4$

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Answer:

Correct Answer: C

Solution:

[c] $\int_{E_{p} d S = \frac{Q + Q^{'}}{ε_{0}}}$

Where Q’ is the charge outside the sphere

$Q_{1}^{'} \int_{R}^{r} d V = \int_{R}^{r} \frac{α}{r} \times 4 π r^{2} d r$

$= 4 π α {(\frac{r^{2}}{2})}^{r} = 4 π α (\frac{r^{2}}{2} - \frac{R^{2}}{2}) = 2 π α (r^{2} - R^{2})$

$E_{p} 4 π r^{2} = \frac{Q + 2 π α (r^{2} - R^{2})}{ε_{0}}$

$E_{p} = \frac{Q}{4 π r^{2} ε_{0}} + \frac{α}{2 ε_{0}} - \frac{α R^{2}}{2 r^{2} ε_{0}}$

E is independent of r if

$\frac{Q}{4 π r^{2} ε_{0}} - \frac{α R^{2}}{2 r^{2} ε_{0}} = 0$

$Q = 2 π R^{2} α$

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Electrostatics Question 736

Options:

Answer:

Solution:

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Electrostatics Question 736

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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