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The electric field is perpendicular to equipotential surface. If it's not the case then it follows that a component can be drawn parallel to the surface which would mean there exists a potential difference. This violates the bare definition of equipotential surface.
(The reason why it's a violation of conservation of energy might be due to the fact that when there is an electric field, there is a force and hence work could be done, which is as you know is not the case with equipotential surface.)
Now, if two equipotential surfaces intersect, it would mean there are two different directions of electric field line which is not possible. Hence your statement follows.
equipotential surface cant intersect because then there will be two direction of electric field lines which in not possible as electric field is a vector quantity and vector are unidirectional.
a parallel plate capacitor has a fixed charge 'q' on its plates. the plates are now pulled further apart, some work must be done during this pull. why? does the potential difference change during the process?
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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
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to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
The electric field is perpendicular to equipotential surface. If it's not the case then it follows that a component can be drawn parallel to the surface which would mean there exists a potential difference. This violates the bare definition of equipotential surface.
(The reason why it's a violation of conservation of energy might be due to the fact that when there is an electric field, there is a force and hence work could be done, which is as you know is not the case with equipotential surface.)
Now, if two equipotential surfaces intersect, it would mean there are two different directions of electric field line which is not possible. Hence your statement follows.
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Isn't that just circular reasoning.
equipotential surface cant intersect because then there will be two direction of electric field lines which in not possible as electric field is a vector quantity and vector are unidirectional.
a parallel plate capacitor has a fixed charge 'q' on its plates. the plates are now pulled further apart, some work must be done during this pull. why? does the potential difference change during the process?