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35.(C) a parabola (D) an ellipseThe electric field at (30, 30) cm due to acharge of -8 nC at the origin in NC is(A) -400(i+1) (B) 400 (i+j)(C)200,2 (i+] (D) 200/2 (1+1)Techarges 4 x 10°C and -16 x 10°C |
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Answer» correct Option C Distance of the charge from the origin = d= The electric field strength is in the direction of Ф where Tan Ф = 30 cm/ 30 cm = 1 => 45 deg. with X- axis. Cos Ф = 1/√2 and Sin Ф = 1/√2Since E is negative, the Ф is in the 3th quadrant. ie., Ф = 180+45 deg. = 225 deg The vector form of electric field = |
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