Physics Exam > Physics Questions > A grounded conducting sphere of radius a is p...
Start Learning for Free
A grounded conducting sphere of radius a is placed with its centre at the origin. A point dipole of dipole moment p=pk is placed at a distance d along the -axis, where 1,k are the units vector along the x and z axes respectively. This leads to the formation of an image dipole of strength p at a distance d' from the centre along the -axis. If d'=a^2/d, then p'=?
Most Upvoted Answer
A grounded conducting sphere of radius a is placed with its centre at ...
Formation of Image Dipole
When a point dipole is placed near a grounded conducting sphere, it induces charges on the surface of the sphere. These induced charges rearrange themselves to form an image dipole. The image dipole has the same magnitude as the original dipole but is located at a specific distance and in a specific orientation from the centre of the sphere.
Position of the Image Dipole
In this case, the original dipole is placed at a distance d along the -axis. The image dipole is formed at a distance d' from the centre along the -axis. The relationship between d and d' is given as d' = a^2/d, where a is the radius of the conducting sphere.
Explanation of the Relationship
To understand why d' = a^2/d, let's consider the electric field produced by the original dipole at the surface of the conducting sphere. The electric field lines produced by the dipole terminate on the surface of the sphere, causing a redistribution of charges.
Electric Field Due to the Original Dipole
The electric field produced by the original dipole at any point in space is given by:
E = (1/4πε₀) * (p/r^3 - 3(p·r)r/r^5)
Where p is the dipole moment, r is the position vector, and ε₀ is the permittivity of free space.
Electric Field at the Surface of the Sphere
At the surface of the conducting sphere, the electric field is perpendicular to the surface. This means that the dot product (p·r) in the equation for the electric field becomes zero. Therefore, the electric field simplifies to:
E = (1/4πε₀) * (p/r^3)
Charge Induced on the Surface
The electric field at the surface of the conducting sphere induces charges on the surface. These induced charges redistribute themselves to form an image dipole.
The charge induced on the surface of the conducting sphere is given by:
q = -σA
Where σ is the surface charge density and A is the area of the surface.
Relationship between Charge and Electric Field
The relationship between the charge induced on the surface and the electric field at the surface is given by Gauss's law:
q = ε₀E
Combining these equations, we can write:
-σA = ε₀ * (1/4πε₀) * (p/r^3)
Simplifying, we get:
σ = -p/(4πr^3)
Formation of the Image Dipole
The induced charges on the conducting sphere redistribute themselves to form an image dipole. The image dipole has the same magnitude as the original dipole but is located at a distance d' from the centre along the -axis.
The distance d' can be found by considering the geometry of the situation. Since the conducting sphere is grounded, the potential at the surface of the sphere is zero. The potential due to the original dipole and the image dipole at the surface of the sphere must cancel each other out.
The potential due to a dipole at a point P in space is given by:
V = (1/4πε₀) * (p·r)/r^3
At the surface of the sphere
When a point dipole is placed near a grounded conducting sphere, it induces charges on the surface of the sphere. These induced charges rearrange themselves to form an image dipole. The image dipole has the same magnitude as the original dipole but is located at a specific distance and in a specific orientation from the centre of the sphere.
Position of the Image Dipole
In this case, the original dipole is placed at a distance d along the -axis. The image dipole is formed at a distance d' from the centre along the -axis. The relationship between d and d' is given as d' = a^2/d, where a is the radius of the conducting sphere.
Explanation of the Relationship
To understand why d' = a^2/d, let's consider the electric field produced by the original dipole at the surface of the conducting sphere. The electric field lines produced by the dipole terminate on the surface of the sphere, causing a redistribution of charges.
Electric Field Due to the Original Dipole
The electric field produced by the original dipole at any point in space is given by:
E = (1/4πε₀) * (p/r^3 - 3(p·r)r/r^5)
Where p is the dipole moment, r is the position vector, and ε₀ is the permittivity of free space.
Electric Field at the Surface of the Sphere
At the surface of the conducting sphere, the electric field is perpendicular to the surface. This means that the dot product (p·r) in the equation for the electric field becomes zero. Therefore, the electric field simplifies to:
E = (1/4πε₀) * (p/r^3)
Charge Induced on the Surface
The electric field at the surface of the conducting sphere induces charges on the surface. These induced charges redistribute themselves to form an image dipole.
The charge induced on the surface of the conducting sphere is given by:
q = -σA
Where σ is the surface charge density and A is the area of the surface.
Relationship between Charge and Electric Field
The relationship between the charge induced on the surface and the electric field at the surface is given by Gauss's law:
q = ε₀E
Combining these equations, we can write:
-σA = ε₀ * (1/4πε₀) * (p/r^3)
Simplifying, we get:
σ = -p/(4πr^3)
Formation of the Image Dipole
The induced charges on the conducting sphere redistribute themselves to form an image dipole. The image dipole has the same magnitude as the original dipole but is located at a distance d' from the centre along the -axis.
The distance d' can be found by considering the geometry of the situation. Since the conducting sphere is grounded, the potential at the surface of the sphere is zero. The potential due to the original dipole and the image dipole at the surface of the sphere must cancel each other out.
The potential due to a dipole at a point P in space is given by:
V = (1/4πε₀) * (p·r)/r^3
At the surface of the sphere
|
Explore Courses for Physics exam
|
|
Similar Physics Doubts
A grounded conducting sphere of radius a is placed with its centre at the origin. A point dipole of dipole moment p=pk is placed at a distance d along the -axis, where 1,k are the units vector along the x and z axes respectively. This leads to the formation of an image dipole of strength p at a distance d' from the centre along the -axis. If d'=a^2/d, then p'=?
Question Description
A grounded conducting sphere of radius a is placed with its centre at the origin. A point dipole of dipole moment p=pk is placed at a distance d along the -axis, where 1,k are the units vector along the x and z axes respectively. This leads to the formation of an image dipole of strength p at a distance d' from the centre along the -axis. If d'=a^2/d, then p'=? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A grounded conducting sphere of radius a is placed with its centre at the origin. A point dipole of dipole moment p=pk is placed at a distance d along the -axis, where 1,k are the units vector along the x and z axes respectively. This leads to the formation of an image dipole of strength p at a distance d' from the centre along the -axis. If d'=a^2/d, then p'=? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A grounded conducting sphere of radius a is placed with its centre at the origin. A point dipole of dipole moment p=pk is placed at a distance d along the -axis, where 1,k are the units vector along the x and z axes respectively. This leads to the formation of an image dipole of strength p at a distance d' from the centre along the -axis. If d'=a^2/d, then p'=?.
A grounded conducting sphere of radius a is placed with its centre at the origin. A point dipole of dipole moment p=pk is placed at a distance d along the -axis, where 1,k are the units vector along the x and z axes respectively. This leads to the formation of an image dipole of strength p at a distance d' from the centre along the -axis. If d'=a^2/d, then p'=? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A grounded conducting sphere of radius a is placed with its centre at the origin. A point dipole of dipole moment p=pk is placed at a distance d along the -axis, where 1,k are the units vector along the x and z axes respectively. This leads to the formation of an image dipole of strength p at a distance d' from the centre along the -axis. If d'=a^2/d, then p'=? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A grounded conducting sphere of radius a is placed with its centre at the origin. A point dipole of dipole moment p=pk is placed at a distance d along the -axis, where 1,k are the units vector along the x and z axes respectively. This leads to the formation of an image dipole of strength p at a distance d' from the centre along the -axis. If d'=a^2/d, then p'=?.
Solutions for A grounded conducting sphere of radius a is placed with its centre at the origin. A point dipole of dipole moment p=pk is placed at a distance d along the -axis, where 1,k are the units vector along the x and z axes respectively. This leads to the formation of an image dipole of strength p at a distance d' from the centre along the -axis. If d'=a^2/d, then p'=? in English & in Hindi are available as part of our courses for Physics.
Download more important topics, notes, lectures and mock test series for Physics Exam by signing up for free.
Here you can find the meaning of A grounded conducting sphere of radius a is placed with its centre at the origin. A point dipole of dipole moment p=pk is placed at a distance d along the -axis, where 1,k are the units vector along the x and z axes respectively. This leads to the formation of an image dipole of strength p at a distance d' from the centre along the -axis. If d'=a^2/d, then p'=? defined & explained in the simplest way possible. Besides giving the explanation of
A grounded conducting sphere of radius a is placed with its centre at the origin. A point dipole of dipole moment p=pk is placed at a distance d along the -axis, where 1,k are the units vector along the x and z axes respectively. This leads to the formation of an image dipole of strength p at a distance d' from the centre along the -axis. If d'=a^2/d, then p'=?, a detailed solution for A grounded conducting sphere of radius a is placed with its centre at the origin. A point dipole of dipole moment p=pk is placed at a distance d along the -axis, where 1,k are the units vector along the x and z axes respectively. This leads to the formation of an image dipole of strength p at a distance d' from the centre along the -axis. If d'=a^2/d, then p'=? has been provided alongside types of A grounded conducting sphere of radius a is placed with its centre at the origin. A point dipole of dipole moment p=pk is placed at a distance d along the -axis, where 1,k are the units vector along the x and z axes respectively. This leads to the formation of an image dipole of strength p at a distance d' from the centre along the -axis. If d'=a^2/d, then p'=? theory, EduRev gives you an
ample number of questions to practice A grounded conducting sphere of radius a is placed with its centre at the origin. A point dipole of dipole moment p=pk is placed at a distance d along the -axis, where 1,k are the units vector along the x and z axes respectively. This leads to the formation of an image dipole of strength p at a distance d' from the centre along the -axis. If d'=a^2/d, then p'=? tests, examples and also practice Physics tests.
|
|
Explore Courses for Physics 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