Physics - Electrostatics Question with Solution | TestHub

PhysicsElectrostaticsElectric Field & ForceEasy2 minQB

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Charge $2 q$ is uniformly distributed on non conducting shell of radius $r /2$ a point charge $q$ is placed at distance $r$ from the centre of the shell as shown. Find the electric field intensity at point P in given diagram.

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Solution:

Electric field intensity = Electric field due to non-conducting shell + Electric field due to $q$ .

$= 0 + \frac{k q}{r ^{2}} (w h ere, r = r^{2} + d^{2})$

$= \frac{k q}{( r ^{2} + d ^{2} ) ^{2}} = \frac{k q}{r ^{2} + d ^{2}}$

Stream:JEESubject:PhysicsTopic:ElectrostaticsSubtopic:Electric Field & Force

⏱ 2mℹ️ Source: QB

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Physics - Electrostatics Question with Solution | TestHub

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