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Byju's Answer

Standard XII

Physics

Polarization Vector

Two dielectri...

Question

Two dielectric slabs of constant $K_{1}$ and $K_{2}$ have been filled in between the plates of a capacitance as shown. What will be the capacitance of the capacitor?

\frac{ε_{0} A}{d} (K_{1} + K_{2})

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\frac{2 ε_{0} A}{d} (\frac{K_{1} + K_{2}}{K_{1} \times K_{2}})

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\frac{d}{2 ε_{0} A}

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\frac{2 ε_{0} A}{d} (\frac{K_{1} \times K_{2}}{K_{1} + K_{2}})

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Solution

The correct option is A $\frac{ε_{0} A}{d} (K_{1} + K_{2})$
capacitance with dielectrics has been introduced capacitors

$C = k \frac{ε_{0} A}{d}$

$C = C_{1} + C_{2}$

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

= $\frac{ε_{0} A (K_{1} + K_{2})}{d}$

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Two dielectric slabs of constant K1 and K2 have been filled in between the plates of a capacitance as shown. What will be the capacitance of the capacitor?

The correct option is A ε0Ad(K1+K2)capacitance with dielectrics has been introduced capacitors C=kε0AdC=C1+C2 = ε0(A/2)K1d+ε0(A/2)K2d = ε0A(K1+K2)d

Two dielectric slabs of constant $K_{1}$ and $K_{2}$ have been filled in between the plates of a capacitance as shown. What will be the capacitance of the capacitor?

The correct option is A $\frac{ε_{0} A}{d} (K_{1} + K_{2})$
capacitance with dielectrics has been introduced capacitors

$C = k \frac{ε_{0} A}{d}$

$C = C_{1} + C_{2}$

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

= $\frac{ε_{0} A (K_{1} + K_{2})}{d}$