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

Question

A charge of -5.00nC is spread uniformly over the surface of one face of a nonconducting disk of radius 1.10cm .

A) Find the magnitude of the electric field this disk produces at a point P on the axis of the disk a distance of 3.00cm 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 and direction of the electric field at point P.

D)Why is the field in part (a) stronger than the field in part (b)?

E)Why is the field in part (c) the strongest of the three fields?

Explanation / Answer

a)E=2pi*k*sigma*(1-x/sqrt(x^2+R^2))=2pi*9*10^9*(-5*10^-9/(pi*0.011^2))*(1-0.03/sqrt(0.03^2+0.011^2))=-45464N/C

b)E=kQx/(x^2+R^2)^1.5=9*10^9*(-5*10^-9)*0.03/(0.03^2+0.011^2)^1.5=-41380N/C

c)E=kq/x^2=9*10^9*-5*10^-9/0.03^2=-50000N/C

d)The field in a) is stronger than that in b) because as we move away from the centre of the ring, there appears a component perpendicular to the axis which is cancelled out. So, for a disc, the net field is sort of an average of the field at as if the charge were at the centre as well as that on the ring. Therefore, the field is stronger than due to the ring and weaker than that at the centre.

e) No component is getting cancelled out. Also, the distance to the point is minimum for the centre of the disc. Both points contribute to the higher field.

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