- AIncreases
- BDecreases
- CRemains constant
- DBecomes zero
Explanation:
When the dielectric slab is introduced between the plates, the induced surface charge on the dielectric reduces the electric field.
The reduction in the electric field results in a decrease in potential difference.
$\text{V}=\text{Ed}=\frac{\text{E}_0\text{d}}{\text{k}}=\frac{\text{V}_0}{\text{k}}$
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