Question

The capacitance of a parallel plate capacitor is 50 pF and the distance between the plates is 4mm. It is charged to 200 V and then the charging battery is removed. Now a dielectric slab (kappa=4) of thickness 2mm is placed. Determine (i) final charge on each plate (ii) finial potential difference between the plates (iii) final energy is the capacitor.

Answer 1

Answer:

The capacitance of a parallel plate capacitor is 50 pF and the distance between the plates is 4mm. It is charged to 200 V and then the charging battery is removed. Now a dielectric slab (kappa=4) of thickness 2mm is placed. Determine (i) final charge on each plate (ii) finial potential difference between the plates (iii) final energy is the capacitor.

Answer 2

Given:

• Capacitance of a parallel plate $= 50 pF$

• Distance between the plates $(d) = 4mm$

• Voltage $(V) = 200V$

• Thickness of slab $(t) = 2 mm$

• Dielectric constant $(k) = 4$

To Find:

(i) final charge on each plate

(ii) finial potential difference between the plates

(iii) final energy is the capacitor

Solution:

Capacitance is given by

     $C = \frac{\epsilon \ A}{d}$

$50 \ pF = \frac{\epsilon \ A}{4}$

When dielectric slab constant is introduced

Capacitance is given by

   $C_ 1 = \frac{ \epsilon A}{\frac{t}{k} + d - t }$

  $C_ 1 = \frac{ \epsilon A}{\frac{2}{4} + 4 - 2 }$

putting the value of $\frac{\epsilon \ A}{4}$

   $C_ 1 = \frac{ 4}{2.5 } \frac{A \epsilon}{4}$

   $C_ 1 = \frac{ 4}{2.5 } \times 50 pF$  

  $C_1 = 80 pF$

Now,

(i) final charge on each plate

        $Q = CV$

        $Q = 50 \times 10^{-12} \times 100$

        $Q = 50 \times 10^{-9}$

        $Q = 50 nC$

Thus, the final charge  $Q = 50 nC$

(ii) finial potential difference between the plates

After charging the battery we removed the battery so the charge will remains same

        $C_1 = 80 pF$

          $V = \frac{Q}{C_1}$

         $V = \frac{ 5 \times 10^{-9}}{80 \times 10^{-12}}$

         $V =\frac{5}{80} \times 10^3 V$

         $V = 62.5 V$

(iii) final energy is the capacitor

         $U = \frac{1}{2} \times C_1 V^2$

         $U = \frac{1}{2} \times 80 \times 10^{-12} \times 62.5 \times 62.5$

         $U = 1.56 \times 10^{-7}J$

Thus,  Final energy of the capacitor  $U = 1.56 \times 10^{-7}J$