Question

A charge of -8.50nC is spread uniformly over the surface of one face of a nonconducting disk of radius 1.00cm .

Part A

Find the magnitude of the electric field this disk produces at a point P on the axis of the disk a distance of 2.50cm from its center.

ANSWER:

E =

9.61?105

N/C

Incorrect; Try Again; 3 attempts remaining

Part 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.

ANSWER:

E =

? N/C

Part 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.

ANSWER:

E = ?

N/C

Part D

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

ANSWER:

Essay answers are limited to about 500 words (3800 characters maximum, including spaces).

3785 Character(s) remaining

Part E

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

ANSWER:

Essay answers are limited to about 500 words (3800 characters maximum, including spaces).

3785 Character(s) remaining

Answer 1

A similar question is solved below, but with different values. Please workout using your figures. Hope this helps you. Please rate me.

A charge of -6.50 nC is spread uniformly over the surface of one face of a nonconducting disk of radius 1.25cm.

A) Find the magnutide and direction of the electric field this disk produces at a point P on the axis of the disk a distance of 2.00 cm from its center.

B) Suppose that the charge were all pushed away from the center and distribute uniformly on the outer rim of the disk. Find the magnitude and direction 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) strnger than the field in part (B)? Why is the field in part (C) the strongest of the three fields?

Charge on disc = Q = - 8*10^ -9 C

radius of disc = R=1.30 cm = 0.013 m

Charge density on disc = d =Q /pi*R^2

Charge density on disc = d = -8*10^-9 / 3.1416*0.013^2

Charge density on disc = d = - 1.5068*10^-5 C/m^2

Electric field due to disc =Ed = - [1.5068*10^-5/1.77*10-11][1-0.838]

Ed= 1.375*10^5 N/C
______________________________

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*8*10^-9*0.02 / [4*10^-4+1.69*10^-4]^3/2

Ering =1.44/ 1.357*10^-5

Ering =1.06*10^5 N/C
___________________------
3-)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*8*10^-9/4*10^-4

E point =1.8*10^5 N/C towards the charge or the direction of Epoint is along negative x axis direction
__________________________________
4) 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