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A thin uniformly charged disc of radius R has surface charge density sigma. At its centre the field is sigma/(2*epsilon0). Compared with this central value, the axial electric field at a distance sqrt(3)*R from the centre is reduced by what percentage?

1. reduces by 70.7%
2. reduces by 29.3%
3. reduces by 86.6%
4. reduces by 13.4%

Correct answer: reduces by 86.6%

Solution

The axial field of a uniformly charged disc is E(x) = (sigma/(2*epsilon0))*(1 - x/sqrt(x²+R²)). At x = sqrt(3)R, sqrt(x²+R²) = sqrt(3R²+R²) = 2R, so x/sqrt(x²+R²) = sqrt(3)R/2R = sqrt(3)/2 ~ 0.866. Thus E = E_centre*(1 - 0.866) = 0.134*E_centre. So the field is reduced to 13.4% of the centre value, i.e. a reduction of 86.6%.

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