A very small uniformly charged plastic ball is placed on top of another in a test tube, as the figure shows: http://bildr.no/view/1194610The balls are in an equilibrium situation with a distance d from each other. If the charge on each ball is doubled, the distance between the balls in the test tube will be;A) √(2d)B) 2dC) 4dD) 8d
1- State your team and Record- Lakers (42-10) #2- What is a positive thing they did the first Half- Played defense very well in the first 7 games. #3. What can they Improve on for the Second Half- Play defense the rest of the way #4- Which Player has been the Most important and most consistent- Kobe and Pau. #5- Which Player is critical for your Team Success in the Second Half- Lamar Odom , Bynum is gonna be out. Odom needs to step up as he did the past couple weeks. The tough loss part the most heartbreaking one was the Troy Murphy tip in, I mean, yeah the Bobcats loss was bad, but you could already tell Lakers were gonna lose since Kobe had fouled out, but here, we are up one.. first guy misses the shot!! We win, and then BAM! a tip in. That just hurts.
Without seeing your picture, it is clear the two are oppositely charged and constrained to move linearly. If the charge is doubled, and uniformly distributed, it is as if the force between them goes up 4 times, so distance must also go up the same to have the same equilibrium ( wrt their weights, I suppose). So your answer is C)
The balls are in an equilibrium situation with a distance d from each other. If the charge on each ball is doubled, the distance between the balls in the test tube will be; The top ball is suspended in the air, so the net force is 0 N. This means the electrostatic force is equal to the weight of the ball. Fs = m* g Fs = k * q1 * q2 ÷ d^2 k * q1 * q2 ÷ d^2 = m * g d^2 = (k * q1 * q2) ÷ (m * g) d^2 = (q1 * q2) * [k ÷ (m * g)] d = √(q1 * q2) * √[k ÷ (m * g)] √[k ÷ (m * g)] is a constant. So, the distance between the 2 balls is directly proportional to the square root of the product of the charges. Let’s double the charges and see what happens to d d = √(2q1 * 2q2) * √[k ÷ (m * g)] d = √(4 * q1 * q2) * √[k ÷ (m * g)] d = 2 * √(q1 * q2) * √[k ÷ (m * g)] Doubling the charge, doubles the distance! the answer is B = 2 * d
