1. Two point charges are separated by 6 cm. The attractive force between them is

Question

1.       Two point charges are separated by 6 cm. The attractive force between them is 20 N. Find the force between them when they are separated by 12 cm. (Why can you solve this problem without knowing the magnitudes of the charges?) <?xml:namespace prefix = o ns = "urn:schemas-microsoft-com:office:office" />

2.       Explain the difference between electrical potential and electrical potential energy.

3.       What is the voltage at the location of a 0.0001-coulomb charge that has an electric potential energy of 0.5 joules?

Explanation / Answer

1) Use the formula F = kq1q2/(r^2) If force is 20N when it is separated by 6 cm, then force will be reduced in quarter when it is separated by 12 cm (6x2) because of (r^2) in denominator 2) Electrical potential (difference) is voltage and electrical potential energy is work done by charge of voltage. 3) P = VI -----> V = P/I = 0.5J/0.0001C = 5000V

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