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A small metal sphere, carrying a net charge `q_1 = -2 mu C`, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of `q_2 = -8 mu C` and mass `1.50 g`, is projected toward `q_1`. When the two spheres are `0.800 m` apart, `q_2` is moving toward `q_1` with speed `20 ms^-1` as shown in (Fig. 3.151). Assume that the two spheres can be treated as point charges. You can ignore the force of gravity.
.
How close does `q_2` get to `q_1` ?

A

0.20 m

B

0.30 m

C

0.10 m

D

0.15 m

Text Solution

Verified by Experts

The correct Answer is:
B

`E_i = K_i + U_i = (1)/(2) (0.0015 kg)(20.0 ms^-1)^2+(k (2.00 xx 10^-6 C)(8.00 xx 10^-6 C))/(0.800 m)`
=`0.48 J`
`E_i = E_f = (1)/(2) mv_f^2 + (k q_1 q_2)/(r_f)`
`v_f = sqrt((2(0.48 J - 0.3t6 J))/(0.0015 kg)) = 4 sqrt(10) ms^-1`
At the closest point, the velocity is zero.
`0.48 J = (kq_1 q_2)/(r)`
or `r = (k(2.00 xx 10^-6 C)(8.00 xx 10^-6 C))/(0.48 J) =0.30 m`.
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