Question

wo parallel-plate capacitors C1 and C2 are connected in parallel to a 12.0 V battery. Both capacitors have the same plate area of 5.30 cm2 and plate separation of 2.65 mm. However, the first capacitor C1 is filled with air, while the second capacitor C2 is filled with a dielectric that has a dielectric constant of 2.10. (a) What is the charge stored on each capacitor

Answer 1

Complete Question

Two parallel-plate capacitors C1 and C2 are connected in parallel to a 12.0 V battery. Both capacitors have the same plate area of 5.30 cm2 and plate separation of 2.65 mm. However, the first capacitor C1 is filled with air, while the second capacitor C2 is filled with a dielectric that has a dielectric constant of 2.10.

(a) What is the charge stored on each capacitor

 (b)  What is the total charge stored in the parallel combination?

Answer:

a

   i    $Q_1 = 2.124 *10^{-11} \ C$

   ii    $Q_2 = 4.4604 *10^{-11} \ C$

b

  $Q_{eq} = 6.5844 *10^{-11} \ C$

Explanation:

From the question we are told that

   The  voltage of the battery is  $V = 12.0 \ V$

    The  plate area of each capacitor is  $A = 5.30 \ cm^2 = 5.30 *10^{-4} \ m^2$

    The  separation between the plates is  $d = 2.65 \ mm = 2.65 *10^{-3} \ m$

     The permittivity of free space  has a value  $\epsilon_o = 8.85 *10^{-12} \ F/m$

     The  dielectric constant of the other material is  $z = 2.10$

The  capacitance of the  first capacitor is mathematically represented as

       $C_1 = \frac{\epsilon * A }{d }$

substituting values

        $C_1 = \frac{8.85 *10^{-12 } * 5.30 *10^{-4} }{2.65 *10^{-3} }$

       $C_1 = 1.77 *10^{-12} \ F$

The  charge stored in the first capacitor is  

       $Q_1 = C_1 * V$

substituting values

        $Q_1 = 1.77 *10^{-12} * 12$

       $Q_1 = 2.124 *10^{-11} \ C$

The capacitance of the second  capacitor is mathematically represented as

       $C_2 = \frac{ z * \epsilon * A }{d }$

substituting values

       $C_1 = \frac{ 2.10 *8.85 *10^{-12 } * 5.30 *10^{-4} }{2.65 *10^{-3} }$

       $C_1 = 3.717 *10^{-12} \ F$

The  charge stored in the second capacitor is  

      $Q_2 = C_2 * V$

substituting values

     $Q_2 = 3.717*10^{-12} * 12$

     $Q_2 = 4.4604 *10^{-11} \ C$

Now  the total charge stored in the parallel combination is mathematically represented as

     $Q_{eq} = Q_1 + Q_2$

substituting values

    $Q_{eq} = 4.4604 *10^{-11} + 2.124*10^{-11}$

     $Q_{eq} = 6.5844 *10^{-11} \ C$