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प्रश्न
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.
विकल्प
a, b and c
a, c and d
b, c and d
c and d
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उत्तर
a, b and c
Explanation:
The density of the equipotential lines gives an idea about the magnitude of electric field. Higher the density, larger the field strength.
We know, the electric field intensity E and electric potential V are related as a, b and c
We know that on any two points of equipotential surface, potential difference is zero or of equal potential.
∵ `E = (-dV)/(dr)`
So the electric field intensity is inversely proportional to the separation between equipotential surfaces.
So equipotential surfaces are closer in regions of large electric. Thus, it verifies answer a
The electric field is larger near the sharp edge, due to larger charge density as a is very small
∵ `sigma = q/A`
So equipotential surfaces are closer or crowded. It verifies answer b.
As the electric field `E = (kq)/r^2` and potential or field decreases as size of the body increases or vice-versa (case of the earth), so the equipotential surfaces will be more crowded if the charge density `sigma = q/A` increases. It verifies the answer c.
As the equipotential surface depends on distance r by `E = (-dV)/r` and `V = (kq)/r`. Equipotential surface depends on charge density at that place which is different at a different place, so equipotential surface are not equispaced all over.
Hence the electric field intensity E is inversely proportional to the separation between equipotential surfaces. So, equipotential surfaces are closer in regions of large electric fields. As electric field intensities is large near sharp edges of charged conductor and near regions of large charge densities. Therefore, equipotential surfaces are closer at such places.
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(i)![]() |
(ii)![]() |
(iii)![]() |
(iv)![]() |
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