Question

A 0.106-A current is charging a capacitor that has square plates 4.60 cm on each side. The plate separation is 4.00 mm.

Answer 1

Answer:

a

 $\frac{d \phi_{E}}{dt} =1.1977 *10^{10} \ V\cdot m/s$

b

 $I = 0.106 \ A$

Explanation:

From the question we are told that

  The current is  $I = 0.106 \ A$

   The length of one side of the square $a = 4.60 \ cm = 0.046 \ m$

    The separation between the plate is  $d = 4.0 mm = 0.004 \ m$

Generally electric flux is mathematically represented as

       $\phi_E = \frac{Q}{\epsilon_o}$

differentiating both sides with respect to t is  

       $\frac{d \phi_{E}}{dt} = \frac{1}{\epsilon_o} * \frac{d Q}{ dt}$

=>     $\frac{d \phi_{E}}{dt} = \frac{1}{\epsilon_o} *I$

Here $\epsilon_o$ is the permitivity of free space with value  

        $\epsilon _o = 8.85*10^{-12} C/(V \cdot m)$

=>   $\frac{d \phi_{E}}{dt} = \frac{0.106}{8.85*10^{-12}}$

=>   $\frac{d \phi_{E}}{dt} =1.1977 *10^{10} \ V\cdot m/s$

Generally the displacement current between the plates in A

    $I = 8.85*10^{-12} * 1.1977 *10^{10}$

=>  $I = 0.106 \ A$