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jok3333 [9.3K]
3 years ago
11

Two large thin metal plates are parallel and close to each other. On their inner faces, the plates have excess surface charge of

opposite signs. The amount of charge per unit area is given by |σ| = 6.8 ✕ 10-22 C/m2, with the negatively charged plate on the left. What are the magnitude and direction of the electric field vector E in unit-vector notation at the following locations?
Physics
1 answer:
wariber [46]3 years ago
5 0

Answer:

For left = 0  N/C

For right = 0  N/C

At middle = -7.6836 * 10^{-11} \vec{i}  N/C

Explanation:

Given data :-

б =6.8 * 10^{-22} C/ m²

Considering the two thin metal plates to be non conducting sheets of charges.

Electric field is given by

E = \frac{\sigma }{2\varepsilon }

1) To the left of the plate

\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+  (\frac{\sigma }{2\varepsilon })(\vec{i})   = 0 N/C.

2) To the right of them.

\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+  (\frac{\sigma }{2\varepsilon })(\vec{i})   = 0 N/C.

3) Between them.

\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+  (\frac{\sigma }{2\varepsilon })(-\vec{i}) = (\frac{\sigma }{\varepsilon })(-\vec{i}) = -\frac{6.8 * 10^{-22} }{8.85 * 10 ^{-12} }  \vec{i} =   -7.6836 * 10^{-11} \vec{i} N/C

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3 years ago
This diagram of the first right-hand rule relates which two quantities?
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At the top of a pole vault, an athlete actually can do work pushing on the pole before releasing it. Suppose the pushing force t
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Answer:

The work done on the athlete is approximately 2.09 J

Explanation:

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\displaystyle{\int_{x_i}^{x_j}F(x)dx}

and substituting with the function of our problem:

\displaystyle{\int_{0}^{0.19}(140x-190x^2)dx\approx2.09\mathrm{J}}

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4 years ago
A Carnot engine operates between temperature levels of 600 K and 300 K. It drives a Carnot refrigerator, which provides cooling
KATRIN_1 [288]

Explanation:

Formula for maximum efficiency of a Carnot refrigerator is as follows.

      \frac{W}{Q_{H_{1}}} = \frac{T_{H_{1}} - T_{C_{1}}}{T_{H_{1}}} ..... (1)

And, formula for maximum efficiency of Carnot refrigerator is as follows.

     \frac{W}{Q_{C_{2}}} = \frac{T_{H_{2}} - T_{C_{2}}}{T_{C_{2}}} ...... (2)

Now, equating both equations (1) and (2) as follows.

 Q_{C_{2}} \frac{T_{H_{2}} - T_{C_{2}}}{T_{C_{2}}} = Q_{H_{1}} \frac{T_{H_{1}} - T_{C_{1}}}{T_{H_{1}}}        

        \gamma = \frac{Q_{C_{2}}}{Q_{H_{1}}}

                    = \frac{T_{C_{2}}}{T_{H_{1}}} (\frac{T_{H_{1}} - T_{C_{1}}}{T_{H_{2}} - T_{C_{2}}})

                    = \frac{250}{600} (\frac{(600 - 300)K}{300 K - 250 K})

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4 0
3 years ago
An open rectangular tank 1 m wide and 2 m long contains gasoline to a depth of 1 m. If the height of the tank sides is 1.5 m, wh
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Answer:

a_y = 4.9\ m/s^2

Explanation:

Given,

Width of rectangular tank, b = 1 m

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Depth of gasoline on the tank, h = 1 m

\dfrac{dz}{dy}=-\dfrac{1.5-1}{1}

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The differential form with the acceleration

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Hence, Horizontal acceleration of the gasoline before gasoline would spill is equal to 4.9 m/s²

3 0
3 years ago
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