$A$ capacitor with plate separation $d$ is charged to $V$ volts. The battery is disconnected and a dielectric slab of thickness $\frac{d}{2}$ and dielectric constant $K=2$ is inserted between the plates. The potential difference across its terminals becomes

  • A
    $V$
  • B
    $2 V$
  • C
    $\frac{4 V}{3}$
  • D
    $\frac{3 V}{4}$

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