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snow_lady [41]
3 years ago
13

A square metal plate of edge length 6.4 cm and negligible thickness has a total charge of 6.6 x 10-6 C. Estimate the magnitude E

of the electric field just off the center of the plate (at, say, a distance of 0.71 mm from the center) by assuming that the charge is spread uniformly over the two faces of the plate. Express your answer in terms of N/C.
Physics
1 answer:
marysya [2.9K]3 years ago
4 0

Answer:

E=9.1\times 10^7 N/C

Explanation:

We are given that

Length of side,l=6.4 cm=6.4\times 10^{-2} m

1 m=100 cm

Total charge,q=6.6\times 10^{-6} C

We have to find the magnitude E of the electric field just off the center of the plate.

Area charge density,\sigma=\frac{q}{A}=\frac{q}{l^2}=\frac{6.6\times 10^{-6}}{(6.4\times 10^{-2})^2}

\sigma=16.1\times 10^{-4}C/m^2

Electric field,E=\frac{\sigma}{2\epsilon_0}

Where \epsilon_0=8.85\times 10^{-12}

Substitute the values

E=\frac{16.1\times 10^{-4}}{2\times 8.85\times 10^{-12}}

E=9.1\times 10^7 N/C

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An aluminum bar 600mm long, with diameter 40mm, has a hole drilled in the center of the bar. The hole is 40mm in diameter and 10
s2008m [1.1K]

Answer:

<em>1.228 x </em>10^{-6}<em> mm </em>

<em></em>

Explanation:

diameter of aluminium bar D = 40 mm  

diameter of hole d = 30 mm

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modulus of elasticity E = 85 GN/m^2  = 85 x 10^{9} Pa

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true length of bar = 600 - 100 = 500 mm

area of the bar A = \frac{\pi D^{2} }{4} =  \frac{3.142* 40^{2} }{4} = 1256.8 mm^2

area of hole a = \frac{\pi(D^{2} - d^{2}) }{4} = \frac{3.142*(40^{2} - 30^{2})}{4} = 549.85 mm^2

Total contraction of the bar = \frac{F*L}{AE} + \frac{Fl}{aE}

total contraction = \frac{F}{E} * (\frac{L}{A} +\frac{l}{a})

==> \frac{180*10^{3}}{85*10^{9}} *( \frac{500}{1256.8} + \frac{100}{549.85}) = <em>1.228 x </em>10^{-6}<em> mm </em>

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