Given-

Capacitance C=5 μF = F

Voltage, V=6v

Separation between plates, d=2 mm=2×10^-3 m

a)The charge on the positive plate is calculated using

q = c×v

where, c is the capacitance

and v = voltage applied

Thus,

q=5 μF×6 V

=30 μC

b) The electric field between the plates of the capacitor is given by

Where, v is the applied voltage and d is the distance between the capacitor plates

⇒

=3×10³ V/m

c)Given-

Distance between the plates of the capacitor, d =2×10^-3 m
Dielectric constant of the dielectric material inserted, k = 5
Thickness of the dielectric material inserted, t = 1×10^-3 m

capacitance of the capacitor= 5 μF

Now,
To calculate area of the plates of the capacitor,

Where,

A = area

k = dielectric constant of the material placed

d= separation between the plates

substituting the values,

When the dielectric placed in it, the capacitance becomes

t=thickness of the material

substituting the values,

1)

d)
The charge stored in the capacitor initially is -

C=5×10^-6 F

Also, V=6 V

Now, we know

where v is the applied voltage and c is the capacitance

⇒

⇒

Now, when the dielectric slab is inserted ,charge on the capacitor, from 1)

Charge, Q’

Now, charge flow is given by,