InterviewSolution
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InterviewSolution
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The capacitance of a capacitor formed by two parallel metal plates each 200 cm2 in area separated by a dielectric 4 mm thick is 0.0004 micro farads. A p.d. of 20,000 V is applied. Calculate (a) the total charge on the plates (b) the potential gradient in V/m (c) relative permittivity of the dielectric (d) the electric flux density. |
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Answer» C = 4 × 10−4 μF ; V = 2 × 104 V (a) ∴ Total charge Q = CV = 4 × 10−4 × 2 × 104 μC = 8 μC = 8 × 10−6 C (b) Potential gradient = dv/dx = (2 x 104)/(4 x 10-3) = 5 x 106V/m (c) D = Q/A = 8 × 10−6 /200 × 10−4 = 4 × 10−4 C/m2 (d) E = 5 × 106 V/m Since D = ε0 εr E ∴ εr = D/(ε0 x E) = (4 x 10-4)/(8.854 x 10-12 x 5 x 106) = 9 |
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