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FIGURE3 10) A uniform electric field with a magnitude of 6 106 N/C is applied to

ID: 1773057 • Letter: F

Question

FIGURE3 10) A uniform electric field with a magnitude of 6 106 N/C is applied to a cube of edge length 0.1 10) m as shown in Fig. 3. If the direction of the E-field is along the x-axis, what is the electric flux passing through the shaded face of the cube? A) 6 x104 Nm2/C B) 0.6 104 Nm2/c C) 600-104 Nm2/C D) 6000 104 Nm2/C E) 60 104 Nm2/C 11) A negative charge, if free, tries to move A) in the direction of the electric field B) from high potential to low potential. C) away from infinity D) toward infinity. E) from low potential to high potential. 12) An equipotential surface must be 12) A) parallel to the electric field at every point B) oriented 30° with respect to the electric field at every point. C) equal to the electric field at every point. D) oriented 60 with respect to the electric field at every point. E) perpendicular to the electric field at every point. 13) Electric dipoles always consist of two charges that are 13) A) equal in magnitude; both are negative. B) unequal in magnitude; opposite in sign C) equal in magnitude; opposite in sign. D) equal in magnitude; both are positive. E) unequal in magnitude; both are negative.

Explanation / Answer


electric flux = E.A ( dot product of E and A )


E = 6*10^6 i N/C

A = area = 0.1^2 i m^2 0.01 i m^2


electric flux = 6*10^6 i . 0.01 i

electric flux = 6*10^4 Nm^2/C <<<----ANSWER

OPTION (A)

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11)


Electric field is directed from high potential to low potential


negative charge always moves from low potential to high potential


OPTION ( E)

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12)


equipotential surface means potential is same

from equation Va - Vb = -E.ds = -E*ds*costheta


if Va = vb

va - vb = 0


E*ds*costheta = 0


theta = 90


equipotential surface is perpendicular to Electric field


OPTION (E)


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13)


dipole has two charges that are equal in magnitude and has opposite sign

OPTION (C)