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प्रश्न
Find the equation of the equipotentials for an infinite cylinder of radius r0, carrying charge of linear density λ.
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उत्तर
To find the potential at distance r from the line consider the electric field. We note that from symmetry the field lines must be radially outward. Draw a cylindrical Gaussian surface of radius r and length l. Then
`oint E.dS = 1/ε_0 λ1`
Or `E_r.2pirl = 1/ε_0 λ1`
⇒ `E_r = lambda/(2piε_0r)`
Hence, if r0 is the radius,
`V(r) - V(r_0) = - int_(r_0)^r E.dl = λ/(2piε_0)ln r_0/r`
For a given V,
ln `r/r_0 = - (2piε_0)/λ [V(r) - V(r_0)]`
⇒ r = r0e –2πε0Vr0/λe + 2πε0V(r)/λ
The equipotential surfaces are cylinders of radius r = r0e –2πε0[V(r) – V(r0)]/λ
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(3) Can electric field exist tangential to an equipotential surface? Give reason
A particle of mass 'm' having charge 'q' is held at rest in uniform electric field of intensity 'E'. When it is released, the kinetic energy attained by it after covering a distance 'y' will be ______.
Equipotentials at a great distance from a collection of charges whose total sum is not zero are approximately.
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- Equipotential surfaces never intersect each other.
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- cannot be defined as `- int_A^B E.dl`
- must be defined as `- int_A^B E.dl`
- is zero.
- can have a non-zero value.
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A charge \[q_0\] is moved from point A to point B on the same equipotential surface. If the potentials at A and B are \[V_A\] and \[V_B\] respectively, the work done is:
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The direction in which electric potential decreases is:
For an electric dipole (+q, −q), the equipotential surfaces are best described as:
In the derivation proving \[\vec{E}\] is perpendicular to an equipotential surface, a charge \[q_0\] undergoes a small displacement \[d\vec{l}\] along the surface, with \[\vec{E}\] making an angle \[\theta\] with the displacement. Equating the work expressions leads to which conclusion?
