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Figure shows a parallel plate capacitor with plate area `A` and plate separation `d`. A potential difference is being applied between the plates. The battery is then disconnected and a dielectric slab of dieletric constant `K` is placed in between the plates of the capacitor as shown. The electric field in the gaps between the plates and the electric slab will beA. `(epsilon_0AV)/d`B. `V/d`C. `(KV)/d`D. `V/(d-t)` |
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Answer» Correct Answer - B `E_(air)=E_0=V/d` |
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