Two equal, positive charges, q=2.0C are located on thex-axis, one at +0.3m and t

Question

Two equal, positive charges, q=2.0C are located on thex-axis, one at +0.3m and the other at -0.3m. A third chargeQ=4.0C is located on the y-axis at +0.4m. A) Find the magnitude and direction of the resultant(net) force on Q. B) Find the potential at the point (0.0, -0.4m). Two equal, positive charges, q=2.0C are located on thex-axis, one at +0.3m and the other at -0.3m. A third chargeQ=4.0C is located on the y-axis at +0.4m. A) Find the magnitude and direction of the resultant(net) force on Q. B) Find the potential at the point (0.0, -0.4m).

Explanation / Answer

A)the force acting between the charges q at +0.3 m and thethird charge Q is F1 = k * (q * Q/r^2) k = (1/4o) = 9 * 10^9 Nm^2/C^2,q = 2.0C = 2.0 * 10^-6 C,Q = 4.0 C = 4.0 * 10^-6 C and r^2 = ((0.4)^2 + (-0.3)^2) = 0.25 m^2 the force acting between the charges q at -0.3 m and the thirdcharge Q is F2 = k * (q * Q/r^2) the magnitude of the resultant (net) force on Q is F = (Fx^2 + Fy^2)^(1/2) or Fx = F1x + F2x = F1 * cos1 + F2 * cos2 and Fy = F1y + F2y = F1 * sin1 + F2 * sin2 Using trinometry we get tan1 = (0.4 - 0/0 - 0.3) = -1.33 and tan2 = (0.4 - 0/0 - (-0.3)) = 1.33 the direction of the resultant (net) force on Q is tan = (Fy/Fx) or = tan-1(Fy/Fx) B)the poetntial due to charge at +0.3 m is V1 = k * (q/r) q = 2.0 C = 2.0 * 10^-6 C and r = ((0 - 0.3)^2 + (-0.4 -0)^2)^(1/2) = 0.70 m the potential due to charge at -0.3 m is V2 = k * (q/r1) r1 = ((0 - (-0.3))^2 + (-0.4 - 0)^2)^(1/2) = 0.70 m the potential at the point (0.0, -0.4m) V = V1 + V2 or V = k * (q/r) + k * (q/r1) = k * q * ((1/r) + (1/r1)) r1 = ((

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