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A test charge q is made to move in the electric field of a point charge Q along two different closed paths (Figure). First path has sections along and perpendicular to lines of electric field - Physics

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प्रश्न

A test charge q is made to move in the electric field of a point charge Q along two different closed paths (Figure). First path has sections along and perpendicular to lines of electric field. Second path is a rectangular loop of the same area as the first loop. How does the work done compare in the two cases?

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उत्तर

Work done will be zero in both cases.

Explanation: The electrostatic field is conservative, and in this field work done by electric force on the charge in a closed loop is zero. In this question both are closed paths, hence the work done in both cases will be zero.

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अध्याय 2: Electrostatic Potential And Capacitance - MCQ I [पृष्ठ १४]

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एनसीईआरटी एक्झांप्लर Physics [English] Class 12
अध्याय 2 Electrostatic Potential And Capacitance
MCQ I | Q 2.18 | पृष्ठ १४

संबंधित प्रश्न

A spherical conducting shell of inner radius r1 and outer radius r2 has a charge Q.

(a) A charge q is placed at the centre of the shell. What is the surface charge density on the inner and outer surfaces of the shell?

(b) Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but has any irregular shape? Explain.


(a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by

`(vec"E"_2 - vec"E"_1).hat"n" = sigma/in_0`

Where `hat"n"` is a unit vector normal to the surface at a point and σ is the surface charge density at that point. (The direction of `hat"n"` is from side 1 to side 2.) Hence show that just outside a conductor, the electric field is σ `hat"n"/in_0`

(b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another.

[Hint: For (a), use Gauss’s law. For, (b) use the fact that work done by electrostatic field on a closed loop is zero.]


If Coulomb’s law involved 1/r3 dependence (instead of 1/r2), would Gauss’s law be still true?


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Which of the following statement is true?


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