Charges + q and - q are arranged on the X axis as shown. [ k = 4 π ε 0 1 ]

V 0 = k q [ 1 − 2 1 + 4 1 + … ∞ ] = k q [ 1 − ( − 2 1 ) 1 ] = 3 2 k q E 0 = − k q [ 1 − 4 1 + 16 1 − 64 1 + … ] = − k q [ 1 − ( − 4 1 ) 1 ] = − 5 4 k q

Physics - Electrostatics Question with Solution | TestHub

PhysicsElectrostaticsRelation between field & potentialMedium2 minQB

PhysicsMediummultiple choice

Charges + q and - q are arranged on the X axis as shown. $[k = \frac{1}{4 π ε _{0}}]$

Options:(select one or more)

Answer:

A, C

Solution:

$V_{0} = k q [1 - \frac{1}{2} + \frac{1}{4} + \dots \infty]$

$= k q [\frac{1}{1 - ( - \frac{1}{2} )}] = \frac{2 k q}{3}$

$E_{0} = - k q [1 - \frac{1}{4} + \frac{1}{16} - \frac{1}{64} + \dots]$

$= - k q [\frac{1}{1 - ( - \frac{1}{4} )}] = - \frac{4 k q}{5}$

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

⏱ 2mℹ️ Source: QB

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