The surface charge density of a thin charged disc of radius R is . The value of the electric field at the centre of the disc is 2 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:

Step-by-step solution

Electric field intensity at the centre of the disc. E = 2 ₀ (given) Electric field along the axis at any distance x from the centre of the disc E ^ = 2 ₀ (1- x x ^2+ R ^2 ) From question, x = R (radius of disc) aligned E ^ &= 2 ₀ (1- R R ^2+ R ^2 ) &= 2 ₀ ( 2 R - R 2 R ) aligned = 4 14 E % reduction in the value of electric field = (E- 4 14 E ) 100 E = 1000 14 %=70.7 %