A parallel plate capacitor with square plates is filled with four dielectrics of dielectric constants $K_{1}, K_{2}, K_{3}, K_{4}$ arranged as shown in the figure. The effective dielectric constant $K$ will be:

Question

A parallel plate capacitor with square plates is filled with four dielectrics of dielectric constants K_1, K_2, K_3, K_4 arranged as shown in the figure. The effective dielectric constant K will be:

MARKS Content Team · Accepted Answer

Correct answer: (A) K=K_1K_2/(K_1+K_2)+K_3.K_4/(K_3+K_4). C_1 and C_2 are in series. The equivalent capacitance of C_1 and C_2 is C_12=(∈_0A)/2d/2K_1+d/2K_2=(∈_0A)/dK_1K_2/(K_1+K_2) C_(3 )and C_(4 )are in series C_34=(∈_0A)/2d/2K_3+d/2K_4=(∈_0A)/dK_3K_4/(K_3+K_4) C_eq=C_12+C_34=(∈_0A)/dK_1K_2/(K_1+K_2)+K_3K_4/(K_3+K_4) ∴K_eq=K_1K_2/(K_1+K_2)+K_3.K_4/(K_3+K_4)

A parallel plate capacitor with square plates is filled with four dielectrics of dielectric constants $K_{1}, K_{2}, K_{3}, K_{4}$ arranged as shown in the figure. The effective dielectric constant $K$ will be:

Solution

Relevant Questions

A parallel plate capacitor with square plates is filled with four dielectrics of dielectric constants K1, K2, K3, K4 arranged as shown in the figure. The effective dielectric constant K will be:

Solution

Relevant Questions

A parallel plate capacitor with square plates is filled with four dielectrics of dielectric constants $K_{1}, K_{2}, K_{3}, K_{4}$ arranged as shown in the figure. The effective dielectric constant $K$ will be: