Two dielectric slabs of dielectric constants k₁ and k₂ are filled in between the two plates, each of area A, of the parallel plate capacitor as shown in the figure. Find the net capacitance of the capacitor.

Solution

Capacitance when dielectric is introduced in between the capacitors is given by

C = K \frac{ε_{\circ} A}{d}

The arrangement shown in the figure is equivalent to a parallel combination of two capacitors, each with plate area A/2 and separation between the plates as d.
Thus, the net capacitance is given by

C = C_{1} + C_{2}

= \frac{ε_{0} (A / 2) k_{1}}{d} + \frac{ε_{0} (A / 2) k_{2}}{d} = \frac{ε_{0} A (k_{1} + k_{2})}{2 d}

For Daily Free Study Material Join wiredfaculty

Electrostatic Potential And Capacitance

Two dielectric slabs of dielectric constants k₁ and k₂ are filled in between the two plates, each of area A, of the parallel plate capacitor as shown in the figure. Find the net capacitance of the capacitor.

Some More Questions From Electrostatic Potential and Capacitance Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Loaction

For Daily Free Study Material Join wiredfaculty

Electrostatic Potential And Capacitance

Two dielectric slabs of dielectric constants k1 and k2 are filled in between the two plates, each of area A, of the parallel plate capacitor as shown in the figure. Find the net capacitance of the capacitor.

Some More Questions From Electrostatic Potential and Capacitance Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Two dielectric slabs of dielectric constants k₁ and k₂ are filled in between the two plates, each of area A, of the parallel plate capacitor as shown in the figure. Find the net capacitance of the capacitor.