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Stels [109]
3 years ago
15

The electric potential V in the space between two flat parallel plates 1 and 2 is given (in volts) by V = 2200x2, where x (in me

ters) is the perpendicular distance from plate 1. At x = 1.7 cm, (a) what is the magnitude of the electric field and (b) is the field directed toward or away from plate 1?
Physics
1 answer:
Bezzdna [24]3 years ago
6 0

Answer:

a) 74.8 V/m

b) direction -(i)

Explanation:

For this case, the electric field is defined as:

E=(\frac{dV}{dx})

because V is a function dependent of x so we can derivate it:

\frac{dV}{dx}=\frac{d(2200x^2)}{dx}=2*2200*x

so:

E=-4400x\\E=-4400V/m^2*0.017m=-74.8V/m

Because of the sign, the magnitude of electric field 74.8 V/m on the negative direction of i vector, thus means towards the plate 1.

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A steel wire of length 31.0 m and a copper wire of length 17.0 m, both with 1.00-mm diameters, are connected end to end and stre
Brut [27]

Answer:

The time taken is  t =  0.356 \ s

Explanation:

From the question we are told that

  The length of steel the wire is  l_1  = 31.0 \ m

   The  length of the  copper wire is  l_2  = 17.0 \ m

    The  diameter of the wire is  d =  1.00 \ m  =  1.0 *10^{-3} \ m

     The  tension is  T  =  122 \ N

     

The time taken by the transverse wave to travel the length of the two wire is mathematically represented as

              t  =  t_s  +  t_c

Where  t_s is the time taken to transverse the steel wire which is mathematically represented as

         t_s  = l_1 *  [ \sqrt{ \frac{\rho * \pi *  d^2 }{ 4 *  T} } ]

here  \rho_s is the density of steel with a value  \rho_s  =  8920 \ kg/m^3

   So

      t_s  = 31 *  [ \sqrt{ \frac{8920 * 3.142*  (1*10^{-3})^2 }{ 4 *  122} } ]

      t_s  = 0.235 \ s

 And

        t_c is the time taken to transverse the copper wire which is mathematically represented as

      t_c  = l_2 *  [ \sqrt{ \frac{\rho_c * \pi *  d^2 }{ 4 *  T} } ]

here  \rho_c is the density of steel with a value  \rho_s  =  7860 \ kg/m^3

 So

      t_c  = 17 *  [ \sqrt{ \frac{7860 * 3.142*  (1*10^{-3})^2 }{ 4 *  122} } ]

      t_c  =0.121

So  

   t  = t_c  + t_s

    t =  0.121 + 0.235

    t =  0.356 \ s

4 0
3 years ago
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