A C D B 3a. Consider the electric charge arrangement of the accompanying figure

Question

A C D B 3a. Consider the electric charge arrangement of the accompanying figure with Q1 +1 nc and Q2 2 nC. Suppose Q1 is located at ac 0 m and Q2 is at z2 1 m. (a) Find the electric field EP (magnitude and direction) half way between them at P. Suppose point A is at 2A 0.40 m and B is at ra 0.60 m. Determine the potentials (b) Vi and (c) Va. (d) Use this information to calculate Suppose point c is at rc 0.49 m and D is at rD 0.51 m. Determine the potentials (e) Vo and (f) VD. (g) Use this information to calculate ac

Explanation / Answer

a) Field at P due to Q1
E(p1) = k Q1 / r2   where r is distance of P from Q1. This field is away from +ve charge Q1, hence along +ve x-axis
E(p1) = 9x109 x 10-9 /0.52 = 36 N/C
similarly E(p2) = 9x109 x 2 x 109 / 0.52 = 72 N/C again along +ve x-axis as Q2 is -ve and towards right of P
Hence E(p) = E(p1) + E(p2) = 108 N/C

b) V(A) = k (Q1/r1 + Q2/r2) where r1 and r2 are distances of A from Q1 qnd Q2 respectively
hence V(A) = 9x109 ( 10-9/0.4 - (2x10-9 ) / 0.6 ) = -7.5 V

c) V(B) = 9x109 ( 10-9/0.6 - (2x10-9 ) / 0.4 ) = -30 V

d) E(p) = - (V(B) - V(A) ) / (Xb - Xa)
   = - ( -30 - (-7.5) ) / ( 0.6 - 0.4) = 112.5 N/C

e) V(C) = 9x109 ( 10-9/0.49 - (2x10-9 ) / 0.51 ) = -16.9 V  
f) V(D) = 9x109 ( 10-9/0.51 - (2x10-9 ) / 0.49 ) = -19.1 V

g) E(p) = - (V(D) - V(C) ) / ( Xd - Xc) = 110 N/C

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