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# physics question?

Two identical, negatively charged metal balls of 1.5 g each hang like pendula from the same nail on strings of length 5.5 cm. They oscillate back and forth, moving together and apart. The strings each have an angle of 11 degrees from the vertical when they are closest together and 15 degrees from the vertical when they are farthest apart. Estimate the number of excess electrons on the metal balls. (k = 9.0 × 10^9 N·m2 /C2 , e = 1.6 x 10^-19 C, me = 9.11 x 10^-31 kg)

### 1 Answer

- NCSLv 71 month ago
I'd go about it this way. The balls exchange GPE for EPE as they oscillate back and forth. The change in height is

h = 0.055m * [(1-cos15º) - (1-cos11º)] = 8.636e-4 m

so the change in gravitational potential energy is

GPE = mgh = 0.0015kg * 9.8m/s² * 8.636e-4m = 1.27e-5 J

The horizontal separation changes from a maximum of

X = 2*0.055m*sin15º = 0.02847 m

to a minimum of

x = 2*0.055m*sin11º = 0.02099 m

Since EPE = kQq/d, and Q=q and k=9.0e9N·m²/C², we get

1.27e-5 J = 9.0e9N·m²/C² * Q² * (1/0.02099m - 1/0.02847m)

which solves to

Q = q = 1.06e-8 C = 10.6 nC

which means the number of excess electrons on each ball is around 6.64e10 (66.4 billion)

Hope this helps!

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