A wheel of diameter 30.0 cm starts from rest and rotates with a constant angular acceleration of 3.50 rad/s2 . At the instant th
e wheel has completed its second revolution, compute the radial acceleration of a point on the rim in two ways.
1 answer:
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Answer:
a) -1.325 μC
b) 4.17μC
Explanation:
First, you need to know that charge is conserved. So, the adition of the charges, as there is no lost in charge, should always be the same. Also, after the wire is removed, both spheres will have the same charge, trying to find equilibrium. In summary:
We know both q1f and q2f must be positive, because the negative charge at the beginning was the the smaller.
The electrostatic force is equal to:
K is the Coulomb constant, equal to 9*10^9 Nm^2/C^2
Now, we are told that the electrostatic force after the wire is equal to 0.0443 N:
Originally, the force is negative because it was an attraction force, therefore, its direction was opposite to the direction of the repulsive force after the wire:
Solving the quadratic equation:
for this values q_1 wil be:
So as you can see, the negative charge will always be -1.325 μC and the positive 4.17μC