σ is the uniform charge density present on the surface of a semi-sphere of radius R. Derive the formula for the electric potential at the centre.
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The given answer is mg/q.
When the oil drop is falling freely due to the effect of gravity at various medium with terminal speed v, then it is given as mg = 6πηrv.
To move the oil drop upward with terminal velocity v if E is the electric field intensity applied, then Eq = mg + 6πηrv = mg + mg = 2mg.
So, E = 2 mg/q.
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