n air gapped parallel plate capacitor has plate separation 0.01 m.(a) (2 marks) The capacitor is charged to 50 V, and remains connectedto the voltage source so that voltage remains at 50 V as a dielectricwith r = 3.4 is inserted so that it completely fills the gap between theplates. Find the magnitudes of the E-field, D-field, and P-field. Ignoreany possible fringing effects.(b) (2 marks) Start again with the capacitor discharged and air gapped.The capacitor is charged to 50 V, but this time the terminals discon-nected, leaving it charged. After the dielectric is inserted, find themagnitudes of the P-field, E-field, and D-field, and the new voltagebetween the plates. Ignore any possible fringing effects
Question
n air gapped parallel plate capacitor has plate separation 0.01 m.(a) (2 marks) The capacitor is charged to 50 V, and remains connectedto the voltage source so that voltage remains at 50 V as a dielectricwith r = 3.4 is inserted so that it completely fills the gap between theplates. Find the magnitudes of the E-field, D-field, and P-field. Ignoreany possible fringing effects.(b) (2 marks) Start again with the capacitor discharged and air gapped.The capacitor is charged to 50 V, but this time the terminals discon-nected, leaving it charged. After the dielectric is inserted, find themagnitudes of the P-field, E-field, and D-field, and the new voltagebetween the plates. Ignore any possible fringing effects
Solution
(a) The electric field (E-field) in a capacitor is given by the equation E = V/d, where V is the voltage and d is the distance between the plates. In this case, E = 50V / 0.01m = 5000 V/m.
The displacement field (D-field) in a dielectric material is given by the equation D = εr * ε0 * E, where εr is the relative permittivity of the material and ε0 is the permittivity of free space (8.85 x 10^-12 F/m). In this case, D = 3.4 * 8.85 x 10^-12 F/m * 5000 V/m = 1.51 x 10^-7 C/m^2.
The polarization field (P-field) in a dielectric material is given by the equation P = ε0 * (εr - 1) * E. In this case, P = 8.85 x 10^-12 F/m * (3.4 - 1) * 5000 V/m = 1.06 x 10^-7 C/m^2.
(b) If the capacitor is disconnected before the dielectric is inserted, the charge on the plates remains constant. This means that the D-field remains constant. Therefore, D = 1.51 x 10^-7 C/m^2.
The E-field in the dielectric is given by the equation E = D / (εr * ε0). In this case, E = 1.51 x 10^-7 C/m^2 / (3.4 * 8.85 x 10^-12 F/m) = 5000 V/m.
The P-field is given by the same equation as in part (a), so P = 1.06 x 10^-7 C/m^2.
The voltage across the plates is given by the equation V = E * d. In this case, V = 5000 V/m * 0.01m = 50V.
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