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 ' $2$ ' is inserted between the plates. The potential difference across its terminals becomes
  • A$V$
  • B$2 V$
  • C$\frac{4 V}{3}$
  • D$\frac{3 V}{4}$
Medium
art

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