Q5/ A circular ring of radius a (meter) has a non-uniform line charge density pL = 2 cos Ø C/m on its perimeter. The ring is located at the z-0 plane, where its center is located at the origin. Find the electric field intensity at: a) The centre of the ring. b) The point (0,0,4).

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Q5/ A circular ring of radius a (meter) has a non-uniform line charge density pL = 2 cos Ø C/m on its
perimeter. The ring is located at the z-0 plane, where its center is located at the origin. Find the electric field
intensity at:
a) The centre of the ring.
b) The point (0,0,4).
Transcribed Image Text:Q5/ A circular ring of radius a (meter) has a non-uniform line charge density pL = 2 cos Ø C/m on its perimeter. The ring is located at the z-0 plane, where its center is located at the origin. Find the electric field intensity at: a) The centre of the ring. b) The point (0,0,4).
Expert Solution
Step 1

Consider the schematic of the given problem as shown below:

 

Physics homework question answer, step 1, image 1

 

Here, O and P are the origin and the given point a distance d above the origin, respectively. The ring-charge has the radius a. As shown, φ represents the angle with respect to the x-axis and θ is the angle between the distance (r) of P from the charge and the x-axis.

Step 2

a)

The electric field at the origin can be visualized as shown by the following schematic:

 

Physics homework question answer, step 2, image 1

 

Here, the charge polarity of the right half of the ring is positive since cosφ is positive there, while it is negative on the other half.

Step 3

The field due to the charge segment (dq) and another segment exactly opposite to it is exactly equal since cos(πφ) is – cosφ.

Due to the symmetry of the cosine function about the x-axis (cos(–φ) = cosφ), all the sine components get canceled. Due to the opposite polarity across the y-axis, however, all the cosine components get doubled.

Evaluate E from Coulomb’s formula as follows:

 

E=14πε0dqa2=14πε02-π2π2ρLadφcosφa2=12πaε0-π2π22cos2φ C/mdφ=1π1 m8.85×10-12 F/mφ2+sin2φ4-π/2π/2 C/m=5.65×1010 V/m

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