In a capacitor of capacitance \(20\,\mu\mathrm F\) the distance between the plates is \(2\,\mathrm{mm}\). If a dielectric slab of width \(1\,\mathrm{mm}\) and dielectric constant 2 is inserted between the plates, then the new capacitance will be:
- A\(22\,\mu\mathrm F\)
- B\(26.6\,\mu\mathrm F\)
- C\(52.2\,\mu\mathrm F\)
- D\(13\,\mu\mathrm F\)
✓ Correct answer: B
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