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Question

# 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.

A
Reduces by 70.7%
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B
Reduces by 29.3%
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C
Reduces by 9.7%
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D
Reduces by 14.6%
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Solution

## The correct option is A Reduces by 70.7%Electric field intensity at the centre of the disc.E=σ2ϵ0(given)Electric field along the axis at any distance x from the centre of the discE′=σ2ϵ0(1−x√x2−R2)From question, x=R (radius of disc)∴E′=σ2ϵ0(1−R√R2+R2)=σ2ϵ0(√2R−R√2R)=414E∴ % reduction in the value of electric field=(E−414E)×100E=100014% =70.7%.

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