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Oliga [24]
3 years ago
9

A neutral solid metal sphere of radius 0.1 m is at the origin, polarized by a point charge of 2 × 10−8 C at location m. At locat

ion m, what is the electric field contributed by the polarization charges on the surface of the metal sphere? (Express your answer in vector form.)
Physics
1 answer:
liraira [26]3 years ago
6 0

Answer: E = 1.8 *10 ^{4} N

Explanation: The formulae for intensity of an electric field of a solid metal sphere relative to a point is given below

E =\frac{Kq}{r^{2} } r

where  k=9* 10^{9}N/m^{2}, q=2 *10 ^{-8} c , r = 0.1m r = is the position vector of the charge.

it has been stated in the question that the charge is placed at the center thus it has no position vector.

E=\frac{9 * 10^{9}* 2* 10^{-8}  }{0.1^{2} }\\ =\frac{18* 10^{1} }{0.01} \\=\frac{18* 10^{1} }{1 *10^{-2} } = 1.8*10^{4} N

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A prismatic bar AB of length L and solid circular cross section (diameter d) is loaded by a distributed torque of constant inten
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Answer:

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Explanation:

Given the data in the question, as illustrated in the image below;

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we know that;

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so

τ_{max} = 16T_{max} /πd³

Therefore, the maximum shear stress τ_{max} the bar is 16T_{max} /πd³

(b) Determine the angle of twist between the ends of the bar.

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d\theta = 32tx.dx / πGd⁴

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\theta =  \int\limits^L_0  \, d\theta

\theta =  \int\limits^L_0  \, 32tx.dx / πGd⁴

\theta = 32t / πGd⁴  \int\limits^L_0  \, xdx

\theta = 32t / πGd⁴ [ L²/2]

\theta = 16tL² / πGd⁴  

Therefore,  the angle of twist between the ends of the bar is 16tL² / πGd⁴  

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From the details given in the question, we are told that:

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