WBJEE · Physics · Capacitance
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 as shown in the figure. The capacitor is now given a charge \(Q\). Then
- A electric field in the dielectric-filled region is higher than that in the air-filled region
- B on the two halves of the bottom plate the charge densities are unequal
- C charge on the half of the top plate above the air-filled part is \(\frac{Q}{K+1}\)
- D capacitance of the capacitor shown above is \((1+K) \frac{C_{0}}{2},\) where \(C_{0}\) is the capacitance of the same capacitor with the dielectric removed.
Answer & Solution
Correct Answer
(D) capacitance of the capacitor shown above is \((1+K) \frac{C_{0}}{2},\) where \(C_{0}\) is the capacitance of the same capacitor with the dielectric removed.
Step-by-step Solution
Detailed explanation
We know that \[ C_{1}=\frac{K \varepsilon_{0} A}{2 d}, C_{2}=\frac{\varepsilon_{0} A}{2 d} \]…
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