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Let there be a spherically symmetric charge distribution with charge density varying as up to r = R and ρ (r) = 0 for r > R, where r is the distance from the origin. The electric field at a distance r (r < r)="" from="" the="" origin="" is="" given="" />
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
Let there be a spherically symmetric charge distribution with charge ...
Consider a thin spherical shell of radius x and thickness dx as shown in the figure.
Volume of the shell, dV = 4πx2dx
Let us draw a Gaussian surface of radius r (r < r)="" as="" shown="" in="" the="" figure="" />
Total charge enclosed inside the Gaussian surface,
According to Gauss's law,
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Let there be a spherically symmetric charge distribution with charge density varying as up to r = R and ρ (r) = 0 for r > R, where r is the distance from the origin. The electric field at a distance r (r a)b)c)d)Correct answer is option 'C'. Can you explain this answer?
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Let there be a spherically symmetric charge distribution with charge density varying as up to r = R and ρ (r) = 0 for r > R, where r is the distance from the origin. The electric field at a distance r (r a)b)c)d)Correct answer is option 'C'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Let there be a spherically symmetric charge distribution with charge density varying as up to r = R and ρ (r) = 0 for r > R, where r is the distance from the origin. The electric field at a distance r (r a)b)c)d)Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let there be a spherically symmetric charge distribution with charge density varying as up to r = R and ρ (r) = 0 for r > R, where r is the distance from the origin. The electric field at a distance r (r a)b)c)d)Correct answer is option 'C'. Can you explain this answer?.
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