A) reduces by 70.7%
B) reduces by 29.3%
C) reduces by 9.7%
D) reduces by 14.6%
Correct Answer: A
Solution :
Electric field intensity at the centre of the disc. \[E=\frac{\sigma }{2{{\in }_{0}}}\](given) Electric field along the axis at any distance x from the centre of the disc \[E'=\frac{\sigma }{2{{\in }_{0}}}\left( 1-\frac{x}{\sqrt{{{x}^{2}}+{{R}^{2}}}} \right)\] From question, x = R (radius of disc) \[\therefore \]\[E'=\frac{\sigma }{2{{\in }_{0}}}\left( 1-\frac{x}{\sqrt{{{R}^{2}}+{{R}^{2}}}} \right)\] \[=\frac{\sigma }{2{{\in }_{0}}}\left( \frac{\sqrt{2}R-R}{\sqrt{2}R} \right)\]\[=\frac{4}{14}E\] \[\therefore \] % reduction in the value of electric field
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