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Scorpion4ik [409]
3 years ago
10

A parallel-plate capacitor in air has circular plates of radius 2.8 cm separated by 1.1 mm. Charge is flowing onto the upper pla

te and off the lower plate at a rate of 5 A. Find the time rate of change of the electric field between the plates.
Physics
1 answer:
Hatshy [7]3 years ago
5 0

Answer:

The time rate of change of the electric field between the plates is  \frac{E }{t} =  2.29 *10^{14} \   N \cdot C  \cdot  s^{-1}  

Explanation:

From the question we are told that

    The  radius is  r =  2.8 \ cm  =  0.028 \ m

     The distance of separation is  d =  1.1  \ mm  =  0.0011 \ m

      The  current is  I  =  5 \ A

Generally the electric field generated is mathematically represented as

         E = \frac{q }{ \pi  *  r^2  \epsilon_o  }

Where \epsilon_o is the permitivity of free space with a value

          \epsilon_o  =  8.85*10^{-12 }\   m^{-3} \cdot kg^{-1}\cdot  s^4 \cdot A^2

So the time rate of change of the electric field between the plates is mathematically represented as

        \frac{E }{t} =   \frac{q}{t} *   \frac{1 }{ \pi  *  r^2  \epsilon_o  }

But \frac{q}{t }  =  I

So  

       \frac{E }{t} =   *   \frac{I }{ \pi  *  r^2  \epsilon_o  }

substituting values  

        \frac{E }{t} =   *   \frac{5 }{3.142  *  (0.028)^2 *   8.85 *10^{-12}  }

        \frac{E }{t} =  2.29 *10^{14} \   N \cdot C  \cdot  s^{-1}

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