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Question
Draw the equipotential surfaces due to an electric dipole.
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Solution
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RELATED QUESTIONS
Two charges 2 μC and −2 µC are placed at points A and B 6 cm apart.
- Identify an equipotential surface of the system.
- What is the direction of the electric field at every point on this surface?
The top of the atmosphere is at about 400 kV with respect to the surface of the earth, corresponding to an electric field that decreases with altitude. Near the surface of the earth, the field is about 100 Vm−1. Why then do we not get an electric shock as we step out of our house into the open? (Assume the house to be a steel cage so there is no field inside!)
Depict the equipotential surfaces for a system of two identical positive point charges placed a distance(d) apart?
Depict the equipotential surface due to
(i) an electric dipole,
(ii) two identical positive charges separated by a distance.
The diagrams below show regions of equipotentials.
(i) |
(ii) |
(iii) |
(iv) |
A positive charge is moved from A to B in each diagram.
If the electric field \[\vec{E}\] had a component along (tangent to) an equipotential surface, what contradiction would arise?
For an infinite line charge (infinite cylinder configuration), the equipotential surfaces are:
In a uniform electric field, the equipotential surfaces take the shape of:
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?
