Suppose a small ball has charge -2.6 muC is located at the origin, and a second

Question

Suppose a small ball has charge -2.6 muC is located at the origin, and a second small ball has charge 6.8 muC is located along the x-axis at +6.5 cm.

(a) What is the x-component of the force on the ball at the origin? Be sure to use the correct sign to represent the direction of the component, + for "to the right", - for "to the left".

(b) What is the x-component of the force on the ball located at 6.5 cm?

(c) Suppose that the second ball at 6.5 cm has charge -6.8 muC. In this case, what is the x-component of the force on the ball at the origin?

(d) The ball is small if its diameter is much less than the separation of the two balls. In this case a ball with diameter 1 mm would be considered small. Such a ball made of ordinary materials would have a mass of no more than 0.01 g. What would the magnitude of the gravitational force of the earth be on such a ball? (Compare the electric and gravitational forces and you can see why we usually can ignore gravity when electric forces are present.)

Explanation / Answer

Given,

q1 = -2.6 uC ; q2 = 6.8 uC

distance = 6.5 cm

a)F = k q1q2/r^2

F = 9 x 10^9 x -2.6 x 10^-6 x 6.8 x 10^-6/0.065^2 = -37.66 N

(The negative sign tells that the force is attractive but as its towards the postitive charge placed at +6.5 cm, it should have a positive magnitude as because of the convention stated in the question)

Hence, Fx = F = 37.66 N

b)Fx = 37.66 N

From Newton's second law, magnitude of force will be same.

c)q2 = -6.8

Fx = F = 9 x 10^9 x -2.6 x 10^-6 x -6.8 x 10^-6/0.065^2 = 37.66 N

But as its towards the left,

Fx = -37.66 N

d)Fg = mg

Fg = 0.01 x 10^-3 x 9.81 = 9.81 x 10^-5 N

We see that the magnitude of gravitational force is very very smallar than that of the electric force and hence its not considered when electric forces are in there.

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