Question

An air-filled parallel-plate capacitor is constructed with a plate area of 0.40 m2 and a plate separation of 0.10 mm. It is then charged to a potential difference of 12 V? (ε0 = 8.85 × 10-12 C2 /N ∙ m2 ) (a) How much charge is stored on each of its plates? (b) If the capacitor is then filled with a glass dielectric (K = 5.0), by how much does the charge if it changes at all? The capacitor is not connected to a battery.

Answer 1

Answer:

The charge stored  is  $Q = 4.25 *10^{-7 } \ C$

The energy stored is  $E = 2.55*10^{-6} \ J$

Explanation:

From the question we are told that

    The area of the plates is  $A = 0.40 \ m^2$

     The separation between the plate is $d = 0.10 \ mm =0.0001 \ m$

      The potential difference is  $V = 12 \ V$

       The permitivity of free space is  $\epsilon_o = 8.85 *10^{-12} C^2 \cdot N^{-1} \cdot m^2$

         The dielectric constant of glass is K =  5.0

Generally the capacitance of this capacitor is

      $C = \frac{\epsilon_o * A}{d}$

substituting values

       $C = \frac{8.85*10^{-12} * 0.40}{0.0001}$

        $C = 3.34 *10^{-8} \ C$

The charge stored is mathematically evaluated as

         $Q = CV$

substituting values

        $Q = (3.54*10^{-8} * 12)$

         $Q = 4.25 *10^{-7 } \ C$

The energy stored is  

         $E = 0.5 * CV^2$

substituting values

         $E = 0.5 * (4.25 *10^{-7} * 12^2)$

           $E = 2.55*10^{-6} \ J$