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What is the field in the cavity if a conductor having a cavity is charged? Does the result depend on the shape and size of cavity or conductor ?

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To solve the question regarding the electric field in the cavity of a charged conductor, we can follow these steps: ### Step-by-Step Solution: 1. **Understanding the Concept of Conductors in Electrostatics**: - When a conductor is charged, the free charges within the conductor redistribute themselves on the surface of the conductor. This redistribution occurs until electrostatic equilibrium is reached. 2. **Electric Field Inside a Conductor**: ...
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A spherical conductor A contains two spherical cavities as shown in Fig.2.127. The total charge on conductor itself is zero . However , there is a point charge q_(1) at the center of one cavity and q_(2) at the center of the other cavity . Another charge q_(3) is placed at a large distance r from the center of the spherical conductor. The force acting on conductor A will be

An ellipsoidal cavity is carved within a perfect conductor. A positive charge q is placed at the centre of the cavity. The points A and B are on the cavity surface as shown in the figure. Then

A spherical conductor A contains two spherical cavities as shown in Fig.2.127. The total charge on conductor itself is zero . However , there is a point charge q_(1) at the center of one cavity and q_(2) at the center of the other cavity . Another charge q_(3) is placed at a large distance r from the center of the spherical conductor. If q_(1) is displaced from its center slightly (being always inside the same cavity ), then the correct representation of field lines inside the same cavity is

We have a fixed conductor having a cavity inside. There is a point charge q_(1) kept inside the cavity charge appearing at the inner surface of the cavity is q_(2) . Charge appearing at outer surface of conductor is q_(3) and there lies another point charge q_(4) out side the conductor. Whole situation is shown in figure. q_(4) . is held stationary by applying an external force vecF_(0) on it. Three points A, B and C are shown. Answer the following two questions. Q. If keeping the magnitude and nature of q_(1) fixed position of q_(1) is changed inside the cavity. Then choose the correct option(s)

A : The field in a cavity inside a conductor is zero which causes electrosatic shielding. R : Dielectric constant of conducotrs in electrostatics is infinite.

A spherical conductor A contains two spherical cavities as shown in Fig.2.127. The total charge on conductor itself is zero . However , there is a point charge q_(1) at the center of one cavity and q_(2) at the center of the other cavity . Another charge q_(3) is placed at a large distance r from the center of the spherical conductor. Which of the following statements are true ?

CENGAGE PHYSICS ENGLISH-ELECTRIC FLUX AND GAUSS LAW-Exercise 2.1
  1. A thin metallic spherical shell contains a charge Q on it. A point cha...

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  2. If we introduce a large thin metal plate between two point charges, wh...

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  3. A dipole lies on the x-axis, with the positive charge +q at x = +d//2 ...

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  4. A Charge Q is distributed uniformly on a ring of radius r. A sphere of...

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  5. Figure shown an imaginary cube of edge L/2. A uniformly Charged rod o...

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  6. A hemispherica body of radius R is placed in a uniform electric field ...

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  7. What is the field in the cavity if a conductor having a cavity is char...

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  8. Figure shows a closed surface which intersects a conducting sphere. If...

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  9. Figure shows a charge q placed at the centre of a hemisphere. A second...

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  10. A charge Q is placed at the centre of an imaginary hemispherical surfa...

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  11. In figure , a cone lies in a uniform electric field E. Determine the e...

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  12. A uniform electric field ahati +bhatj intersects a surface of area A. ...

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  13. A charge Q is placed at a distance a/2 above the centre of a horizonta...

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  14. Calculate the total electric flux through the paraboloidal surface due...

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  15. Consider a closed surface of arbitatry shape as shown in figure. Suppo...

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  16. If Coulomb's law involved 1/r^3 (instead of 1/r^2), would Gauss's law ...

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