JEE Exam > JEE Questions > A sphere of radius R carries charge density p...
Start Learning for Free
A sphere of radius R carries charge density proportional to the square of the distance from the center: ρ = Ar2, where A is a positive constant. At a distance of R/2 from the center, the magnitude of the electric fieldis
- a)A/(4πε0)
- b)AR3/(40ε0)
- c)AR3/(24ε0)
- d)AR3/(5ε0)
Correct answer is option 'B'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Most Upvoted Answer
A sphere of radius R carries charge density proportional to the square...
Understanding the problem:
The charge density of the sphere is given by ρ = Ar^2, where A is a positive constant. We need to find the magnitude of the electric field at a distance of R/2 from the center of the sphere.
Calculating the electric field:
- To find the electric field at a distance of R/2 from the center, we will make use of Gauss's Law.
- Gauss's Law states that the electric field at a distance r from the center of a uniformly charged sphere is given by E = (Qenclosed)/(4πε0r^2), where Qenclosed is the charge enclosed by the Gaussian surface.
- For a sphere of radius R, the charge enclosed within a distance r from the center is Qenclosed = (4/3)πr^3Ar^2 = (4πA/3)r^5.
- Substituting this value of Qenclosed in the expression for the electric field, we get E = (4πA/3)(r^5)/(4πε0r^2) = AR^3/(12ε0)r.
- At a distance of R/2 from the center (i.e., r = R/2), the electric field is E = AR^3/(12ε0)(R/2) = AR^3/(24ε0).
Therefore, the magnitude of the electric field at a distance of R/2 from the center of the sphere is given by option 'b' - AR^3/(24ε0).
The charge density of the sphere is given by ρ = Ar^2, where A is a positive constant. We need to find the magnitude of the electric field at a distance of R/2 from the center of the sphere.
Calculating the electric field:
- To find the electric field at a distance of R/2 from the center, we will make use of Gauss's Law.
- Gauss's Law states that the electric field at a distance r from the center of a uniformly charged sphere is given by E = (Qenclosed)/(4πε0r^2), where Qenclosed is the charge enclosed by the Gaussian surface.
- For a sphere of radius R, the charge enclosed within a distance r from the center is Qenclosed = (4/3)πr^3Ar^2 = (4πA/3)r^5.
- Substituting this value of Qenclosed in the expression for the electric field, we get E = (4πA/3)(r^5)/(4πε0r^2) = AR^3/(12ε0)r.
- At a distance of R/2 from the center (i.e., r = R/2), the electric field is E = AR^3/(12ε0)(R/2) = AR^3/(24ε0).
Therefore, the magnitude of the electric field at a distance of R/2 from the center of the sphere is given by option 'b' - AR^3/(24ε0).
Free Test
FREE
| Start Free Test |
Community Answer
A sphere of radius R carries charge density proportional to the square...
ρ(r) = A(r)2
Charge enclosed for sphere of radius R/2
Charge enclosed for sphere of radius R/2
Applying Gauss’s law for this sphere
Attention JEE Students!
To make sure you are not studying endlessly, EduRev has designed JEE study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in JEE.
|
Explore Courses for JEE exam
|
|
Top Courses for JEEView all
A sphere of radius R carries charge density proportional to the square of the distance from the center: ρ = Ar2, where A is a positive constant. At a distance of R/2 from the center, the magnitude of the electric fieldisa)A/(4πε0)b)AR3/(40ε0)c)AR3/(24ε0)d)AR3/(5ε0)Correct answer is option 'B'. Can you explain this answer?
Question Description
A sphere of radius R carries charge density proportional to the square of the distance from the center: ρ = Ar2, where A is a positive constant. At a distance of R/2 from the center, the magnitude of the electric fieldisa)A/(4πε0)b)AR3/(40ε0)c)AR3/(24ε0)d)AR3/(5ε0)Correct answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A sphere of radius R carries charge density proportional to the square of the distance from the center: ρ = Ar2, where A is a positive constant. At a distance of R/2 from the center, the magnitude of the electric fieldisa)A/(4πε0)b)AR3/(40ε0)c)AR3/(24ε0)d)AR3/(5ε0)Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A sphere of radius R carries charge density proportional to the square of the distance from the center: ρ = Ar2, where A is a positive constant. At a distance of R/2 from the center, the magnitude of the electric fieldisa)A/(4πε0)b)AR3/(40ε0)c)AR3/(24ε0)d)AR3/(5ε0)Correct answer is option 'B'. Can you explain this answer?.
A sphere of radius R carries charge density proportional to the square of the distance from the center: ρ = Ar2, where A is a positive constant. At a distance of R/2 from the center, the magnitude of the electric fieldisa)A/(4πε0)b)AR3/(40ε0)c)AR3/(24ε0)d)AR3/(5ε0)Correct answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A sphere of radius R carries charge density proportional to the square of the distance from the center: ρ = Ar2, where A is a positive constant. At a distance of R/2 from the center, the magnitude of the electric fieldisa)A/(4πε0)b)AR3/(40ε0)c)AR3/(24ε0)d)AR3/(5ε0)Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A sphere of radius R carries charge density proportional to the square of the distance from the center: ρ = Ar2, where A is a positive constant. At a distance of R/2 from the center, the magnitude of the electric fieldisa)A/(4πε0)b)AR3/(40ε0)c)AR3/(24ε0)d)AR3/(5ε0)Correct answer is option 'B'. Can you explain this answer?.
Solutions for A sphere of radius R carries charge density proportional to the square of the distance from the center: ρ = Ar2, where A is a positive constant. At a distance of R/2 from the center, the magnitude of the electric fieldisa)A/(4πε0)b)AR3/(40ε0)c)AR3/(24ε0)d)AR3/(5ε0)Correct answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of A sphere of radius R carries charge density proportional to the square of the distance from the center: ρ = Ar2, where A is a positive constant. At a distance of R/2 from the center, the magnitude of the electric fieldisa)A/(4πε0)b)AR3/(40ε0)c)AR3/(24ε0)d)AR3/(5ε0)Correct answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A sphere of radius R carries charge density proportional to the square of the distance from the center: ρ = Ar2, where A is a positive constant. At a distance of R/2 from the center, the magnitude of the electric fieldisa)A/(4πε0)b)AR3/(40ε0)c)AR3/(24ε0)d)AR3/(5ε0)Correct answer is option 'B'. Can you explain this answer?, a detailed solution for A sphere of radius R carries charge density proportional to the square of the distance from the center: ρ = Ar2, where A is a positive constant. At a distance of R/2 from the center, the magnitude of the electric fieldisa)A/(4πε0)b)AR3/(40ε0)c)AR3/(24ε0)d)AR3/(5ε0)Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of A sphere of radius R carries charge density proportional to the square of the distance from the center: ρ = Ar2, where A is a positive constant. At a distance of R/2 from the center, the magnitude of the electric fieldisa)A/(4πε0)b)AR3/(40ε0)c)AR3/(24ε0)d)AR3/(5ε0)Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A sphere of radius R carries charge density proportional to the square of the distance from the center: ρ = Ar2, where A is a positive constant. At a distance of R/2 from the center, the magnitude of the electric fieldisa)A/(4πε0)b)AR3/(40ε0)c)AR3/(24ε0)d)AR3/(5ε0)Correct answer is option 'B'. Can you explain this answer? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup