Consider a disk of radius 4 cm with a uniformly distributed charge of +5.8 µC. C

Question

Consider a disk of radius 4 cm with a uniformly distributed charge of +5.8 µC. Compute the magnitude of the electric field at a point on the axis and 2.7 mm from the center. The value of the Coulomb constant is 8.98774 × 109 N · m2 /C 2 . Answer in units of N/C. 007 (part 2 of 4) 5.0 points Compute the field from the near-field approximation x R . Answer in units of N/C. 008 (part 3 of 4) 8.0 points Compute the electric field at a point on the axis and 42 cm from the center of the disk. Answer in units of N/C. 009 (part 4 of 4) 5.0 points Approximate the field by treating the disk as a +5.8 µC point charge at a distance of 42 cm. Answer in units of N/C.

Explanation / Answer

Solution: The disk has q = +5.8C charge uniformly distributed on it.

q = +5.8 C*(1C/106 C)

q = +5.8*10-6 C

The radius of disk,

R = 4 cm

R = (4 cm)*(1m/100cm)

R = 0.04 m

Surface area (on one side) of the disk,

A = *R2

A = 3.1416*(0.04 m)2

A = 5.0265*10-3 m2

Thus the charge per unit area is given by,

= q/A

= (+5.8*10-6 C)/(5.0265*10-3 m2)

= 1.1539*10-3 C/m2

The electric field E at a point (which x meters away from the center) along the line passing through the center of disk (of radius R meters) is given by,

E = [/2o]*[1 – x/(x2 + R2)] ------------------------------------------------------------------(1)

Part (1)

We want to find the magnitude of electric field at x = 2.7 mm

x = 2.7 mm*(1m / 1000mm)

x = 2.7*10-3 m

Thus from equation (1) we get,

E = [/2o]*[1 – x/(x2 + R2)]

E = [(1.1539*10-3 C/m2)/(2*8.85*10-12 C2/N.m)]*[1 – (2.7*10-3 m)/( (2.7*10-3 m)2 + (0.04 m)2)]

E = 6.0802*107 N/C

Hence the answer is 6.0802*107 N/C.

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Part (2)

For the same value of x = 2.7 mm, we see that x is very small as compared to the radius of the disk.

The near-field approximation x<<R implies that the term x/(x2 + R2) becomes small and negligible.

Thus the equation reduces to,

E = /2o

The electric field using near-field approximation now becomes,

E = [(1.1539*10-3 C/m2)/(2*8.85*10-12 C2/N.m)]

E = 6.5192*107 N/C

Thus the answer is 6.5192*107 N/C

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Part (3)

In this case, x = 42 cm

x = 42 cm*(1m/100cm)

x = 0.42 m

By putting this value of x in equation (1), we get

E = [/2o]*[1 – x/(x2 + R2)]

E = [(1.1539*10-3 C/m2)/(2*8.85*10-12 C2/N.m)]*[1 – (0.42 m)/( (0.42 m)2 + (0.04 m)2)]

E = 2.9366*105 N/C

Hence the answer is 2.9366*105 N/C

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Part (4)

Since now x>>R, the disk can be treated as the point charge of +5.8C. The electric field at x = 42 cm thus becomes,

E = (1/4o)*q/x2

where (1/4o) = Coulomb constant = 8.98774*109 N·m2/C2

E = (8.98774*109 N·m2/C2)*(5.8*10-6 C)/(0.42m)2

E = 2.9552*105 N/C

Hence the answer is 2.9552*105 N/C

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