Half of the space between the plates of a par | vedclass

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(B, C, D) The capacitor can be viewed as two capacitors in parallel, each with area $A/2$ and plate separation $d$.The capacitance of the dielectric-f

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capacitance of the capacitor shown above is $\frac{(1+K) C_{0}}{2}$, where $C_{0}$ is the capacitance of the same capacitor with the dielectric removed.

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(B, C, D) The capacitor can be viewed as two capacitors in parallel, each with area $A/2$ and plate separation $d$.The capacitance of the dielectric-filled part is $C_1 = \frac{K \varepsilon_0 (A/2)}{d} = \frac{K C_0}{2}$ and the air-filled part is $C_2 = \frac{\varepsilon_0 (A/2)}{d} = \frac{C_0}{2}$, where $C_0 = \frac{\varepsilon_0 A}{d}$.The equivalent capacitance is $C_{eq} = C_1 + C_2 = \frac{C_0}{2}(K+1)$. Thus, option $(d)$ is correct.Since the capacitors are in parallel, the potential difference $V$ across both is the same. The charges are $Q_1 = C_1 V$ and $Q_2 = C_2 V$.The ratio of charges is $\frac{Q_1}{Q_2} = \frac{C_1}{C_2} = K$. Since $Q_1 + Q_2 = Q$, we have $Q_2 = \frac{Q}{K+1}$ and $Q_1 = \frac{KQ}{K+1}$. Thus, option $(c)$ is correct.The charge densities are $\sigma_1 = Q_1 / (A/2)$ and $\sigma_2 = Q_2 / (A/2)$. Since $Q_1 
eq Q_2$, the charge densities are unequal. Thus, option $(b)$ is correct.The electric fields are $E_1 = \frac{\sigma_1}{K \varepsilon_0}$ and $E_2 = \frac{\sigma_2}{\varepsilon_0}$. Substituting $\sigma_1 = K \sigma_2$, we get $E_1 = \frac{K \sigma_2}{K \varepsilon_0} = \frac{\sigma_2}{\varepsilon_0} = E_2$. The fields are equal, so $(a)$ is incorrect.

Half of the space between the plates of a parallel-plate capacitor is filled with a dielectric material of dielectric constant $K$. The remaining half contains air. The capacitor is now given a charge $Q$. Then:

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$A$ capacitor, when filled with a dielectric $K = 3$, has charge $Q_0$, voltage $V_0$, and electric field $E_0$. If the dielectric is replaced with another one having $K = 9$, the new values of charge, voltage, and field will be respectively:

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