Physics - Electrostatics Question with Solution | TestHub

PhysicsElectrostaticsRelation between field & potentialMedium2 minQB

PhysicsMediuminteger

Some equipotential surfaces are shown in figure. What can you say about the magnitude of the electric field (in V/m) ?

Answer:

200

Solution:

Electric field is perpendicular to equipotential surfaces. Adjacent surfaces differ by $Δ V = 10 V$ .

Perpendicular separation: $Δ d = 10 sin 3 0^{\circ} cm = 0.05 m$ .

$∣ E ∣ = \frac{Δ V}{Δ d} = \frac{10}{0.05} = 200 V/m$ . Answer: 200.

Stream:JEESubject:PhysicsTopic:ElectrostaticsSubtopic:Relation between field & potential

⏱ 2mℹ️ Source: QB

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