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प्रश्न
Why is there no work done in moving a charge from one point to another on an equipotential surface?
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उत्तर
On an equipotential surface, the potential remains constant and thus potential difference (ΔV) is zero. The work done on a charge q is given as
W = qΔV
Now, as ΔV = 0
We conclude that W = 0
So, the work done in moving a charge from one point to another on an equipotential surface is zero.
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संबंधित प्रश्न
Draw equipotential surfaces:
(1) in the case of a single point charge and
(2) in a constant electric field in Z-direction. Why are the equipotential surfaces about a single charge not equidistant?
(3) Can electric field exist tangential to an equipotential surface? Give reason
Can two equipotential surfaces intersect each other?
Equipotential surfaces ______.
- are closer in regions of large electric fields compared to regions of lower electric fields.
- will be more crowded near sharp edges of a conductor.
- will be more crowded near regions of large charge densities.
- will always be equally spaced.
Find the equation of the equipotentials for an infinite cylinder of radius r0, carrying charge of linear density λ.
The direction in which electric potential decreases is:
For an infinite line charge (infinite cylinder configuration), the equipotential surfaces are:
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?
A point charge of \[1\ \mu\text{C}\] produces a potential \[V = \frac{kq}{r}\]. At what distance from the charge does the potential equal 90 V? (Take \[k = 9 \times 10^9\ \text{N m}^2/\text{C}^2\])
