A charge of -5.00 nC is spread uniformly over the surface of one face of a nonco

Question

A charge of -5.00 nC is spread uniformly over the surface of one face of a nonconducting disk of radius 1.25 cm . a)Find the magnitude of the electric field this disk produces at a point P on the axis of the disk a distance of 3.00 cm from its center. b)Suppose that the charge were all pushed away from the center and distributed uniformly on the outer rim of the disk. Find the magnitude of the electric field at point P. c)If the charge is all brought to the center of the disk, find the magnitude of the electric field at point P. d)Why is the field in part (a) stronger than the field in part (b)?

Explanation / Answer

a) Charge on disc = Q = - 5*10^-9 C
radius of disc = R=1.25 cm = 0.0125 m
Charge density on disc = d =Q /pi*R^2
Charge density on disc = d = -5*10^-9 / 3.1416*0.0125^2  = - 1.019*10^-5 C/m^2
Electric field due to disc =Ed = = - [1.019*10^-5/1.77*10-11][1-0.923]

Ed= 4.43*10^4 N/C

b) Electric field due to ring = Ering =kQx/[x^2+R^2]^3/2
Ering = kQx / [x^2+R^2]^3/2
Ering =9*10^9*5*10^-9*0.03 / [9*10^-4+1.5625*10^-4]^3/2
Ering =3.93*10^4 N/C

c) If the charge is all brought to the center of the disk, the magnitude and direction of the electric field at point P=Epoint
E point =9*10^9*5*10^-9/9*10^-4
E point =5*10^4 N/C towards the charge or the direction of Epoint is along negative x axis direction

d) The field of disc is stronger than field of the ring because in ring , the charge distribution is farther away as compared to disc where charge near the center is closer
In ring there is no charge at center of ring

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