# Find the electric field between the sheets of the capacitor in the following case after the dielectric plate is introduced.Two sheets of a capacitor have a charge on them equal to 53.1*10^-8 C. The...

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

Two sheets of a capacitor have a charge on them equal to 53.1*10^-8 C. The capacitance is 177 pF and the plates are 1 cm apart. A dielectric plate with a dielectric constant 3 is later introduced between the plates.

justaguide | College Teacher | (Level 2) Distinguished Educator

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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.