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Goryan [66]
3 years ago
6

For the parallel plates mentioned above, the DC power supply is set to 31.5 Volts and the plate on the right is at x = 14 cm. Wh

at is the magnitude of the electric field at a point on the x-axis where x = 7.0 cm? Answer with a number in the format ### in Newtons per Coulombs.

Physics
1 answer:
Aleonysh [2.5K]3 years ago
8 0

Note: The complete question is attached as a file to this solution. The parallel plate mentioned can be seen in this picture attached.

Answer:

E = 225 N/C

Explanation:

Note: At any point on the parallel plates of a capacitor, the electric field is uniform and equal.

Therefore, Electric field at x = 14 cm equals the electric field at x = 7 cm

V(x) = 31.5 Volts

x = 14 cm = 0.14 m

The magnitude of the electric field at any point between the parallel plate of the capacitor is given by the equation:

E = V(x)/d

E(x = 0.14) = 31.5/0.14

E(x=0.14) = 225 N/C

E(x=0.14) = E(x=0.07) = 225 N/C

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Now the general formula of the Doppler's effect is:
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Note: We do not need to worry about the signs, as everything is moving towards each other. If something/somebody were moving away, we would have the negative sign. However, in this problem it is not the issue.

Where,
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v_{r} = Velocity of the receiver/observer relative to the medium = ?.
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Plug-in the above values in the equation (A), you would get:

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3 years ago
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Answer:

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Explanation:

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We need to find the net torque about the origin on a flea located at coordinates. Its formula is given by :

\tau=r\times F

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On solving the cross product of r and F, we get :

\tau=3i+2j+4k

So, the net torque about the origin on a flea is 3i+2j+4k. Hence, this is the required solution.

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