Problem 21.14 Your answer is incorrect. Try again E=E E E1 E = E1 The electric f

Question

Problem 21.14 Your answer is incorrect. Try again E=E E E1 E = E1 The electric field is measured all over a cubical surface, and the pattern of field detected is shown in the figure above. On the right side of the cube, the electric field has magnitude E, 357 V/m, and the angle between the electric field and the surface of the cube is ? = 17 degrees. On the bottom of the cube, the electric field has the same magnitude EI , and the angle between the electric field and the surface of the cube is also ? = 17 degrees. On the top of the cube and the left side of the cube, the electric field is zero. On half of the front and back faces, the electric field has magnitude E1 and is parallel to the face; on the other half of the front and back faces, the electric field is zero. One edge of the cube is 62 cm long. What is the net electric flux on this cubical surface? Net electric flux What is the net charge inside this cubical surface? lc

Explanation / Answer

a) flux = E1 * A * sin 17

Fluxnet = 357 * (0.62 * 0.62) sin 17 + 357 * (0.62 * 0.62) sin 17

net electric flux = 80.24 V.m

b) From gauss law q / e0 = 80.24

q = 8.85 * 10-12 * 80.24

= 7.1 * 10-10 C

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