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

Electric field intensity at the centre of the disc∴E=σ2ϵ01−RR2+R2 ∴reduction in the value electric fieldE=σ2ϵ0( given )∆E=σ2∈02R−R2R=414EPercentage change=E−414E×100E=100014≃71%

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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 $\frac{σ}{2 ϵ_{0}}$ . 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

$∴ E = \frac{σ}{2 ϵ_{0}} (1 - \frac{R}{\sqrt{R^{2} + R^{2}}}) ∴$ reduction in the value electric field

$E = \frac{σ}{2 ϵ_{0}} (given)$

$∆ E = \frac{σ}{2 \in_{0}} (\frac{\sqrt{2} R - R}{\sqrt{2} R}) = \frac{4}{14} E$

$Percentage change = \frac{(E - \frac{4}{14} E) \times 100}{E} = \frac{1000}{14} ≃ 71 %$

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