A parallel-plate of capacitor of capacitance `5mu F` is connected in a battery of emf `6V.` The separation between the plate is 2mm.(a) Find the charge on the positive plate.(b) find the eletric field between the plate.(c) A dielectric slab of thickness 1 mm and dielectric constant 5 is inserted into the gap to occupy the lower half of it.Find the capacitance of the new combination .(d)How much charge has flow n through the battery after the slab is inserted?

Explanation

Let's solve the problem step by step.

### Given:
- Capacitance, \( C = 5 \, \mu F = 5 \times 10^{-6} \, F \)
- EMF of the battery, \( V = 6 \, V \)
- Separation between the plates, \( d = 2 \, mm = 2 \times 10^{-3} \, m \)

### (a) Find the charge on the positive plate.
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