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 10%. Double check your calculations. 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

Given,

R=0.0324 m, Q=5.03*10-6 C,

E=2*3.14*k**(1- x/(R2+x2)0.5)

a) x= 3.03 mm = 3.03*10-3 m = 0.00303 m

=Q/A = 5.03*10-6 / 3.14*0.03242 =3.296*10-3 C/m2

putting all the values in above equation

E=2*3.14*k**(1- x/(R2+x2)0.5)

E=2*3.14*9*109*3.296*10-3*(1- 0.00303/(0.03242+ 0.003032)0.5)

E=1.689*108 N/C away from the center

b) E= /2*0

E=3.296*10-3 / 2*8.85*10-12

E=1.862*108 N/C away from the center

c) x= 0.303 m

E=2*3.14*9*109*3.296*10-3*(1- 0.303/(0.03242+ 0.3032)0.5)

E= 1.055*106 N/C

d) x=0.303 m

For a charged particle, E= k*Q/x2

E=9*109*5.03*10-6 / 0.3032

E=4.930*105 N/C

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