Electrostatics

JEE Main level, single correct, medium

132 of 311

Question 132 of 311

The surface charge density of a thin charged disc of radius R\mathrm{R} is σ\sigma. The value of the electric field at the centre of the disc is σ2∈0\frac{\sigma}{2 \in_0}. With respect to the field at the centre, the electric field along the axis at a distance R\mathrm{R} from the centre of the disc:

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Answer: (A) reduces by 70.7%70.7 \%

Electric field intensity at the centre of the disc.
E=σ2ϵ0 (given) \mathrm{E}=\frac{\sigma}{2 \epsilon_0} \quad \text { (given) }
Electric field along the axis at any distance xx from the centre of the disc
E′=σ2ϵ0(1−xx2+R2)\mathrm{E}^{\prime}=\frac{\sigma}{2 \epsilon_0}\left(1-\frac{\mathrm{x}}{\sqrt{\mathrm{x}^2+\mathrm{R}^2}}\right)
From question, x=R\mathrm{x}=\mathrm{R} (radius of disc)
∴E′=σ2ϵ0(1−RR2+R2)=σ2ϵ0(2R−R2R)\begin{aligned} \therefore \mathrm{E}^{\prime} &=\frac{\sigma}{2 \epsilon_0}\left(1-\frac{\mathrm{R}}{\sqrt{\mathrm{R}^2+\mathrm{R}^2}}\right) \\ &=\frac{\sigma}{2 \epsilon_0}\left(\frac{\sqrt{2} \mathrm{R}-\mathrm{R}}{\sqrt{2} \mathrm{R}}\right) \end{aligned}
=414E=\frac{4}{14} \mathrm{E}
∴%\therefore \% reduction in the value of electric field
=(E−414E)×100E=100014%=70.7%=\frac{\left(E-\frac{4}{14} E\right) \times 100}{E}=\frac{1000}{14} \%=70.7 \%

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