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.
1 Answer
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.