1. Two Ping-Pong balls each have a mass of 2.7 gand carry a net charge of 0.500

Question

1. Two Ping-Pong balls each have a mass of 2.7 gand carry a net charge of 0.500 C . One ball is held fixed. How far apart are they now?

2. Two small charged balls have a repulsive force of 0.20 N when they are separated by a distance of 0.95 m . The balls are moved closer together, until the repulsive force is 0.60 N . At what height (h) should the second ball be placed directly above the fixed ball if it is to remain at rest there? (Here "h" refers to the center to center distance between the two balls.)

3. Coulomb's law for the magnitude of the force Fbetween two particles with charges Q and Qseparated by a distance d is

|F|=K|QQ|d2,

where K=140, and 0=8.854×1012C2/(Nm2) is the permittivity of free space.

Consider two point charges located on the x axis: one charge, q1 = -18.0 nC , is located at x1 = -1.730 m ; the second charge, q2 = 39.5 nC , is at the origin (x=0.0000).

What is the net force exerted by these two charges on a third charge q3 = 54.0 nC placed between q1 and q2 at x3 = -1.150 m ?

4. Three charges are fixed in the xy plane as follows: 1.7 nC at the origin (0, 0); 2.5 nC at (0.75 m , 0); –1.8 nC nC at (0, 1.75 m ). Find the force acting on the charge at the origin.

5. A point charge +6.1 C is placed at the origin, and a second charge –3.6 C is placed in the x-yplane at (0.35 m , 0.55 m ). Where should a third charge be placed in the plane so that the net force acting on it is zero?

Explanation / Answer

1)

Let d be the distance.

F = weight = m g = k q^2/d^2

so

d = sqrt [ k q^2 /(m g) ]

= sqrt [ 9*109 x (0.5*10-6)2 / (0.0027*9.8) ]

= 0.29 m (or 29cm)

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