A small ball of mass, m = 5.10 times 10degree g, has acquired an excess charge.

Question

A small ball of mass, m = 5.10 times 10degree g, has acquired an excess charge. The ball is then placed between two parallel plates spaced x = 3 90 mm which have a potential difference of V = 2010 V applied across them. In this configuration the ball appears to be motion loss or floating in between the plates. What is the overall charge on the ball? Calculate the number of electrons, ng, the ball has either gained or lost. The acceleration due to gravity is g = 9.81 m/s2, and the elementary unit of charge is e = 1.60 times 1019 C. The ball is charged positively neutrally negatively

Explanation / Answer

Potential difference V = 2010 volt Sepration of the plates d = 3.9 x 10 -3 m Mass of the ball m = 5.1 x 10 -9 g                              = 5.1 x 10 -12 kg Electric field E = V / d                         = 515.38 x 10 3 V / m If the weight of the ball is equal to the force on charged particle then it floats. So,mg = Eq From this charge q = mg / E      where g = 9.8m/s 2                               = 9.69 x 10 -17 C   The charge is positive. Number of electrons lost N = q / e Where e = charge of electron                = 1.6 x 10 -19 C Substitute values we get N = 606

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