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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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Complete Question

A commuter train passes a passenger platform at a constant speed of 39.6 m/s. The train horn is sounded at its characteristic frequency of 350 Hz.

(a)

What overall change in frequency is detected by a person on the platform as the train moves from approaching to receding

(b) What wavelength is detected by a person on the platform as the train approaches?

 

Answer:

a

  \Delta  f  =  81.93 \ Hz

b

  \lambda_1 =  0.867 \ m

Explanation:

From the question we are told that

      The speed of the train is  v_t  =  39.6 m/s

      The frequency of the train horn is  f_t =  350 \ Hz

Generally the speed of sound has a constant values of  v_s  =  343 m/s

  Now  according to dopplers equation when the train(source) approaches a person on the platform(observe) then the frequency on the sound observed by the observer can be mathematically represented as  

        f_1 =  f *   \frac{v_s}{v_s - v_t}

substituting values

        f_1 =  350 *  \frac{343 }{343-39.6}

       f_1 =  395.7 \ Hz

  Now  according to dopplers equation when the train(source) moves away from  the  person on the platform(observe) then the frequency on the sound observed by the observer can be mathematically represented as  

           f_2 =  f *   \frac{v_s}{v_s +v_t}

substituting values

        f_2 =  350 *   \frac{343}{343  + 39.6}

       f_2 =  313.77 \ Hz

The overall change in frequency is detected by a person on the platform as the train moves from approaching to receding is mathematically evaluated as

        \Delta  f  =  f_1 - f_2

        \Delta  f  =  395.7 - 313.77

        \Delta  f  =  81.93 \ Hz

Generally the wavelength detected by the person as the train approaches  is mathematically represented  as

          \lambda_1 =  \frac{v}{f_1 }

          \lambda_1 =  \frac{343}{395.7 }

         \lambda_1 =  0.867 \ m

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