Question

Two charges, +5 µC and +17 µC, are fixed 1 m apart, with the second one to the right. Find the magnitude and direction of the net force (in N) on a −7 nC charge when placed at the following locations. (a) halfway between the two magnitude .003024 Correct: Your answer is correct. N direction Correct: Your answer is correct. (b) half a meter to the left of the +5 µC charge magnitude .001736 Correct: Your answer is correct. N direction Correct: Your answer is correct. (c) half a meter above the +17 µC charge in a direction perpendicular to the line joining the two fixed charges (Assume this line is the x-axis with the +x-direction toward the right. Indicate the direction of the force in degrees counterclockwise from the +x-axis.) magnitude N direction ° counterclockwise from the +x-axis †

Answer 1

(c)

Force on -7 nC charge due to 5 uC,

F1 = (kq1q2 / r^2) * (-sin i - cos j)

where, r = sqrt (1² + 0.5²) = 1.118 m

sin = 1 / 1.118

cos = 0.5 / 1.118

F1 = (9*10⁹*5*10^-6*7*10^-9 / 1.118²) * (-1 / 1.118 i - 0.5 / 1.118 j)

F1 = -2.253*10^-4 i - 1.1269*10^-4 j

Force on -7 nC charge due to 17 uC,

F2 = kq2*q3 / r^2 (-j)

F2 = 9*10⁹*17*10^-6*7*10^-9 / 0.5² (-j)

F2 = -4.284*10^-3 (j)

Net force on -7nC,

Fnet = F1 + F2

Fnet = (-2.253*10^-4 i - 1.1269*10^-4 j - 4.284*10^-3 j) = (-2.253*10^-4 i - 4.396*10^-3 j)

Magnitude of resultant force,

F = sqrt [(-2.253*10^-4)² + (-4.396*10^-3)²]

F = 0.0002447 N

Direction,

tan = - 4.396*10^-3 / -2.253*10^-4

= 87.06 deg counterclockwise from the +x-axis