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Q.

The surface charge density of a thin charged disc of radius R is σ. The value of the electric field at the centre of the disc is σ2ϵ0. With respect to the field  at the centre, the electric field along the axis at a distance R from the centre of the disc is reduced by ____ percentage 

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answer is 71.

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Detailed Solution

Electric field intensity at the centre of the disc

E=σ2ϵ01RR2+R2  reduction in the value electric field

E=σ2ϵ0( given )

E=σ202RR2R=414E

Percentage change=E414E×100E=10001471%

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