These question came on a test of mine recently all one after another so it surprises me how the first and second one as so similar (i think)(adsbygoogle = window.adsbygoogle || []).push({});

Determine the electric field on the equipotential surface V=0, and verify that it is perpendicular to this surface at all points on this surface

now since V = 0 , then [itex] \int E ds = 0 [/itex]

So does this mean that the electric field is zero?? Something with zero potential surely has an electric field...

Since V = 0 , then E must be perpendicular to the element of the length of this surface ds, thus E is perpendicular to all s o nthsi surrface.

But is the electric field simply [itex] \frac{q}{\epsilon_{0} A} [/itex]

Where q is the enclosed charge in this surface?

Given that the electric field is conservative i.e. E(r) dr = -dV(r)

Prove that hte electric field lines are perpendicular to the equipotential surfaces at every points on the equipotential surfaces

not quite sure to do here... is nt the answer from the above question the same as this one??

Verify that the surraces of conductors (at equilibrium) are equipotential surfaces

completely clueless on ths ione... does this use the fact taht i = q/t ? But what is meant by the equilibrium state here??

Than kyou for any help and advice you can offer!

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# Electric Field and potential

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