Find the electric field between the sheets of the capacitor in the following case after the dielectric plate is introduced.Two sheets of a...

The plates of a capacitor carry opposite charges. Here it
is given that the charge on the plate is 53.1*10^-8 C. The capacitance of the capacitor
is 177 pF and the sheets are 1 cm apart.

We can calculate
the voltage between the plates using the formula C = Q/V, where Q is the charge on the
plates, V is the voltage between the plates and C is the
capacitance.

As C and Q are known, we can rewrite C =
Q/V

=> V = Q/C

=
53.1*10^-8 C / 177 pF

= 53.1*10^-8 C / 177*10^-12
F

= 3000 V.

Now the
introduction of a dielectric decreases this voltage by a factor equal to the dielectric
constant. So the voltage here decreases to 3000/ 3 =
1000V.

The electric field is equal to Voltage / distance
between the plates = 1000 / (1/100) = 1000*100 = 10^5 V/
m.

The required electric field is equal to
10^5 V/m.