Two identical small metal spheres with q1 > 0 and |q1| > |q2| attract each other with a force of magnitude 72.1 mn when separated by a distance of 1.41 m . f21 f12 1.41 m q1 q2 r1 = 21 µm r2 = 21 µm the spheres are then brought together until they are touching, enabling the spheres to attain the same final charge q. q1 → q q2 → q ∆ q after the charges on the spheres have come to equilibrium, they spheres are separated so that they are again 1.41 m apart. f21 f12 1.41 m q q now the spheres repel each other with a force of magnitude 21.63 mn. what is the final charge on the sphere on the right? the value of the coulomb constant is 8.98755 × 109 n · m2 /c 2 . answer in units of µc. 0what is the initial charge q1 on the first sphere? answer in units of µc.

1)

The electrostatic force between two charges is given by

(1)

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

When the two spheres are brought in contact with each other, the charge equally redistribute among the two spheres, such that each sphere will have a charge of

where Q is the total charge between the two spheres.

So we can actually rewrite the force as

And since we know that

r = 1.41 m (distance between the spheres)

(the sign is positive since the charges repel each other)

We can solve the equation for Q:

So, the final charge on the sphere on the right is

2)

Now we know the total charge initially on the two spheres. Moreover, at the beginning we know that

(we put a negative sign since the force is attractive, which means that the charges have opposite signs)

r = 1.41 m is the separation between the charges

And also,

So we can rewrite eq.(1) as

Solving for q1,

Since , we can substituting all numbers into the equation:

which gives two solutions:

Which correspond to the values of the two charges. Therefore, the initial charge q1 on the first sphere is

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what is the speed of a softball that travels 20m in 2secs

speed = 20/2 = 10 m/s