1. In the figure below, q = 14 ? C and d = 17 cm. In the figure below, q = 14 ?C

Question

1. In the figure below, q = 14 ?C and d = 17 cm.

In the figure below, q = 14 ?C and d = 17 cm. Find the direction and magnitude of the net electrostatic force exerted on the point charge q2 in the figure above. How would your answers to part (a) change if the distance d were increased by a factor of 5? Find the direction and magnitude of the net electrostatic force exerted on the point charge q2 in the figure below. Let q =+1.4 mu C and d = 23 cm. Direction (from the x-axis which points to the right) = Magnitude=

Explanation / Answer

1)

Force due to q1

F1 = k(q)(2q)/d^2 (-i)

F2 = k(2q)(3q)/d^2 (i)

a)

Fa = F1 + F2 = kq^2/d^2 (-2i+6i)

Fa = kq^2 / d^2 *(4i)

Fa = 9*4*10^9*14*14*10^-12/(0.17)^2

Fa = 244.15 N

direction => towards right side

b)

d2 = 5d

Fb = 9*4*10^9*14*14*10^-12/(0.17*5)^2

Fb = 9.766 N

2)

Force due to q1 =>

F1 = k(q)(2q)/d^2 (-j)

F4 = k(4q)(2q)/2d^2 (-i-j)/sqrt(2)

F3 = k(3q)(2q)/d^2 (-i)

Fnet = F1 + F4 + F3

Fnet = kq^2/d^2 ( -j -2sqrt(2)i - 2sqrt(2)j - 6i)

Fnet = 9*10^9*1.4*1.4*10^-12/(0.23)^2

Fnet = -0.33346 (8.828 i + 3.828 j )

Direction =>

tan(theta) = -3.828/-8.828

theta = 23.44 + 180 = 203.44 deg from +ve x-axis

Magnitude =>

|Fnet| = 0.33346 * sqrt(8.828^2 + 3.828^2) = 3.208 N

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