The total electric field at a point on the axis of a uniformly charged disk, whi

Question

The total electric field at a point on the axis of a uniformly charged disk, which has a radius R and a uniform charge density of , is given by the following expression, where x is the distance of the point from the disk.

Ex = 2ke

1

Consider a disk of radius R = 3.24 cm having a uniformly distributed charge of +5.03 C.

(a) Using the expression above, compute the electric field at a point on the axis and 3.03 mm from the center.
  
Your response differs from the correct answer by more than 100%. N/C

(b) Explain how the answer to part (a) compares with the field computed from the near-field approximation

E = /20.

Score: 1 out of 1

Comment:

(c) Using the first expression above, compute the electric field at a point on the axis and 30.3 cm from the center of the disk.
  
Your response differs from the correct answer by more than 100%. N/C

(d) Explain how the answer to part (c) compares with the electric field obtained by treating the disk as a +5.03-C charged particle at a distance of 30.3 cm.

Explanation / Answer

(a)
Area, A = pi*r^2
= pi*0.0324^2
= 3.3* 10^-3 m^2
Q = 5.03 * 10^-6 C
so,
s = Q/A = 1.52* 10^-3 C/m^2

so,
E = 2*pi* (8.99 * 10^9) * (1.52 * 10^-3)* [ 1 - {(3.03 * 10^-3)/((3.24*10^-2)^2 + (3.03*10^-3)^2)^0.5 ]
E = 77.8 * 10^6 N/C

(b)
with near field approximation, we assume,
x<<R
so, in denominator,
R^2 >> x^2
so, x^2 term can be neglected
now fraction in brackets becomes:-
[ 1 - x/R]
as, x<<R,
so, x/R -> 0
so,
E = 2*pi*k*s
as, k = 4*pi*e
so,
E = s/2e

solving:-
E = (1.52*10^-3)/(8.85 * 10^-12)
E = 1.7* 10^8 N/C

so we see that there is very much difference.
the ratio of two Es is :
1.7 * 10^8/(77.8* 10^6)
= 2.2 approx

so, the value we get by near-field approximation is more than twice the value computed without neglecting x.

(c)
E = 2*pi* (8.99 * 10^9) * (1.52 * 10^-3)* [ 1 - {(0.303)/((3.24*10^-2)^2 + ((30.3*10^-2)^2)^0.5 ]
E = 2.96 * 10^5 N/C

(d)
E = k*q/r^2
E = (8.99 * 10^9) * (5.03 * 10^-6) / (0.303)^2
E = 4.9 * 10^5 N/C

so, we see that field is almost same. However field by point charge is slightly greater than field due to chardef disc.
This hapoens because in charged disc, the distance between charge on disc and point on its axis goes on increasing as we move towards circumference of disc from its centre,
and since E is inversely proportional to r^2.

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