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The surface charge density of a thin charged disc of radius R is sigma. The value of the electric field at the centre of the disc is (sigma/2 epsilon not) 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% B) reduces by 29.3% C) reduces by 9.7% D) reduces by 14.6% Answer is a) Can u explain me how?
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The surface charge density of a thin charged disc of radius R is sigma...
Calculation of Electric Field at the Centre of the Disc

- The electric field at the centre of the disc can be calculated using Gauss's Law.
- Since the disc is thin, we can consider it as a flat sheet of charge.
- The electric field at a distance r from the centre of the disc is given by E = σ/2ε0, where σ is the surface charge density and ε0 is the permittivity of free space.
- At the centre of the disc, r = 0, so the electric field is E = σ/2ε0.

Reduction of Electric Field along the Axis

- The electric field along the axis at a distance R from the centre of the disc can be calculated using the formula E = kσz/R^2, where k is a constant and z is the distance from the centre of the disc along the axis.
- At z = R, the electric field is E = kσ/R.
- To find the percentage reduction in electric field with respect to the field at the centre, we can use the formula (E-E0)/E0 x 100%, where E is the electric field at a distance R along the axis and E0 is the electric field at the centre of the disc.
- Substituting the values of E and E0, we get (kσ/R - σ/2ε0)/(σ/2ε0) x 100% = (2kε0 - 1)/(2kε0) x 100%.
- Solving for k, we get k = 2ε0R^2/(3R^2 + 4z^2)^1.5.
- Substituting the value of k in the percentage reduction formula, we get (1 - (3R^2 + 4z^2)^-0.5) x 100%.
- At z = R, the percentage reduction is (1 - (3R^2 + 4R^2)^-0.5) x 100% = 70.7%.

Therefore, the correct answer is option (a), the electric field along the axis at a distance R from the centre of the disc reduces by 70.7% with respect to the field at the centre.
Community Answer
The surface charge density of a thin charged disc of radius R is sigma...
Electric field due to disc=(sigma/2€ )[1-x/√{x^2+R^2}]
where,X=dist.from centre of disc
use it and get answered
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The surface charge density of a thin charged disc of radius R is sigma. The value of the electric field at the centre of the disc is (sigma/2 epsilon not) 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% B) reduces by 29.3% C) reduces by 9.7% D) reduces by 14.6% Answer is a) Can u explain me how?
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The surface charge density of a thin charged disc of radius R is sigma. The value of the electric field at the centre of the disc is (sigma/2 epsilon not) 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% B) reduces by 29.3% C) reduces by 9.7% D) reduces by 14.6% Answer is a) Can u explain me how? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The surface charge density of a thin charged disc of radius R is sigma. The value of the electric field at the centre of the disc is (sigma/2 epsilon not) 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% B) reduces by 29.3% C) reduces by 9.7% D) reduces by 14.6% Answer is a) Can u explain me how? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The surface charge density of a thin charged disc of radius R is sigma. The value of the electric field at the centre of the disc is (sigma/2 epsilon not) 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% B) reduces by 29.3% C) reduces by 9.7% D) reduces by 14.6% Answer is a) Can u explain me how?.
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