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olganol [36]
3 years ago
14

A charge of uniform linear density 2.00 nC/m is distributed along a long, thin, nonconducting rod. The rod is coaxial with a lon

g conducting cylindrical shell (inner radius = 5.60 cm, outer radius = 9.80 cm). The net charge on the shell is zero. (a) What is the magnitude of the electric field at distance r = 15.4 cm from the axis of the shell? What is the surface charge density on the (b) inner and (c) outer surface of the shell?
Physics
1 answer:
NemiM [27]3 years ago
3 0

Answer:

(a) E=233.56 N/C

(b) The surface charge density of inner surface  σ= -5.69×10⁻⁹C/m²

(c)The surface charge density of outer surface  σ= 3.25×10⁻⁹C/m²

Explanation:

For Part (a)

The magnitude of the electric field at distance of 15.4 cm from the axis of the shell is given by:

E=λ/2πε₀r

Substitute the given values

E=\frac{2.00*10^{-9} }{2\pi 8.854*10^{-12}*0.154 }\\ E=233.56N/C

Since the nonconducting rod positively charged,it induces a negative charge -q on the inner surface of conducting shell and a positive charge +q on the outer surface of conducting shell,so the net charge of conducting shell is zero

Part (b)

The surface charge density of inner surface is given by:

σ=-q/A

=\frac{-q}{2\pi r_{inner} l}\\ =\frac{\frac{-q}{l} }{2\pi r_{inner}}\\

= -λ/2πr

=\frac{-2.0*10^{-9} }{2\pi (0.056)}\\ =-5.69*10^{-9}C/m^{2}

Part(c)

Similarly the surface charge density on the outer surface of the cylindrical shell given by:

σ=λ/2πr

=\frac{2.0*10^{-9} }{2\pi 0.098}\\ =3.25*10^{-9}C/m^{2}  

 

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3 years ago
We wrap a light, nonstretching cable around a 10.0 kg kg solid cylinder with diameter of 38.0 cm cm . The cylinder rotates with
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Answer: 14.16

Explanation:

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here we have

J = (3.25 \times 10^8)r^2

now plug in the values in above equation

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