Question

Problem 2 Two infinite parallel planes are separated 022 cm. The planes have equal and opposite charge densities. The charge density of the positive plane is 9.4 C/m Calculate (a) the magnitude of the electric field between the planes. 1.1 MV/m (b) the potential difference from the positive plane to the negative plane. -2.3 kV (c) the potential along the equipotential surface 0.10 cm from the positive plate Let the potential along the positive plane equal 3.0 kV. 19 kV

Answer 1

In the boxes answers of three options is given.

sigma = 9.4 times 1-^-6 e/m^2 (a) electric field due to single plate = sigma/2 epsilon_0 at point P E vector_P = E vector_1 + E vector_2 E_P = sigma/2 epsilon_0 + sigma 2 epsilon_0 E_P = sigma/epsilon_0 vert epsilon_0 = 8.854 times 10^-12 Nm^2/c^2 E_p = 9.4 times 10^-6/8.854 times 10^-12 N/c = 1.06 m v/m E_p = 1.1 M v/m (b) Delta V = -integral_0^0.22 E dr = -integral 1.1 times 10^6 times 22 times 10^-4 V (c) Delta V_1A = -1.06 times 10^6 times 10 times 10^-4 v = -1.06 kv V_A - V_1 = -1.06 kV V_A = -1.06 + 3 kv V_A = 1.9 kV

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