Two charges, Q1= 2.30 ?C, and Q2= 6.30 ?C are located at points (0,-4.00 cm ) an

Question

Two charges, Q1= 2.30 ?C, and Q2= 6.30 ?C are located at points (0,-4.00 cm ) and (0,+4.00 cm), as shown in the figure. What is the magnitude of the electric field at point P, located at (6.00 cm, 0), due to Q1 alone? 3.98

Explanation / Answer

(A) Let we placed a test charge of +q at P, than the Forces on it due to Q1 & Q2 will be same in magnitude and the horizontal component of these will cancel each other and the vertical component of these will add,Let the angle they make from the Y-axis is A*:- =>Fx = 0 as F(Q1) x sinA* = -F(Q2) x sinA* =>Fy = F(Q1) x cosA* + F(Q2) x cosA* =>Fy = 2 x cosA* x kQ1q/d^2 =>Fy = 2 x 1.5/[sqrt{(1.5)^2 + (3)^2}] x 9 x 10^9 x 12 x 10^-3 x q/[sqrt{(1.5)^2 + (3)^2}]^2 =>Fy = 8.59q x 10^6 N----------------------(i) The force on the q by Q2:- =>F(-y) = -kQ2q/(1.5)^2 =>F(-y) = -9 x 10^9 x 0.6 x 10^-3 x q/(2.25) =>F(-y) = - 2.4q x 10^6 N-------------------(ii) Thus F(net) on q = (8.59 - 2.4) x q x 10^6 N =>E at P = F(net)/q = 6.19 x 10^6 N/C in +Y direction --------------------(iii) (B) F = qE =>F = -3.5 x 10^-3 x 6.19 x 10^6 =>F = - 2.17 x 10^4 N [in -Y direction] (C)4. is in the downward direction

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