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Physics - Electrostatics Question with Solution | TestHub

PhysicsElectrostaticsRelation between field & potentialMedium2 minQB

PhysicsMediuminteger

Electric field due to a uniformly charged ring of radius $5 m$ at a certain point on its axis is $1 V / m$ and electric potential at the same point is 6 V. If the distance of this point from centre of the ring is x (in m), find the smallest value of x.

Answer:

Solution:

For a ring, $\frac{E}{V} = \frac{x}{x ^{2} + a ^{2}}$ with $a^{2} = 5$ , so $\frac{x}{x ^{2} + 5} = \frac{1}{6}$ .

$x^{2} - 6 x + 5 = 0$ gives $x = 1$ or $x = 5$ ; the smallest value is $x = 1$ .

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

⏱ 2mℹ️ Source: QB

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