magnitude 1 N/C direction 2° (counterclockwise from the+ x -axis) (a) Three poin

Question

magnitude 1 N/C direction 2° (counterclockwise from the+x-axis) (a) Three point charges, A = 2.2 muC, B = 7.5 muC, and C = - 3.9 muC, are located at the corners of anequilateral triangle as in the figure above. Find the magnitude anddirection of the electric field at the position of the 2.2 muC charge. (b) How would the electric field at that point be affected if thecharge there were doubled? Themagnitude of the field would be halved. Thefield would be unchanged. Themagnitude of the field would double. Themagnitude of the field would quadruple. (c) Would the magnitude of the electric force be affected? no yes

Explanation / Answer

The force acting on the charge at A due to charge at Bis F1 = k * (qAqB/r2)------------(1) k = (1/4o) = 9 * 109Nm2/C2,qA = 2.2 C = 2.2 *10-6 C,qB = 7.5 C = 7.5 *10-6 C and r2 = (0.500)2 = 0.25m2 The force acting on the charge at A due to charge at Cis F2 = k *(qAqC/r12)-------------(2) qC = -3.9 C = -3.9 * 10-6 Cand r12 = (0.500)2 = 0.25m2 The net force acting on the charge at A is Fnet = (Fx2 +Fy2)1/2 -------------(3) Fx = F1x + F2x =F1 * cos1 + F2 *cos2 and Fy = F1y + F2y =F1 * sin1 + F2 *sin2 The angles are 1 = 2 =60o The direction of the net electric force is tan = (Fy/Fx) or =tan-1(Fy/Fx) (counterclockwise from the +x-axis)
The magnitude of the electric field at A is Fnet = E * q or E = (Fnet/q) (b)When the charge is doubled then qA' =2qA qB' = 2qB and qC' = 2qC From equations (1) and (2) we get F1 = k *(qAqB/r2) or F1' = k *(qA'qB'/r2) = k * (2qA* 2qB/r2) = 4k *(qAqB/r2) ------------(4) similarly from equation (2) we get F2 = k *(qAqC/r12) or F2' = 4k *(qAqC/r12)------------(5) Therefore,the net force is four times the intial force.Hencewe can say that the magnitude of the electric field isquadruple. (c)Yes,the magnitude of the electric force will beaffected.The magnitude of the electric force will be fourtimes the initial force magnitude. From equations (1) and (2) we get F1 = k *(qAqB/r2) or F1' = k *(qA'qB'/r2) = k * (2qA* 2qB/r2) = 4k *(qAqB/r2) ------------(4) similarly from equation (2) we get F2 = k *(qAqC/r12) or F2' = 4k *(qAqC/r12)------------(5) Therefore,the net force is four times the intial force.Hencewe can say that the magnitude of the electric field isquadruple. (c)Yes,the magnitude of the electric force will beaffected.The magnitude of the electric force will be fourtimes the initial force magnitude.

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