Here we can use coulomb's law to find the force between two charges
As per coulombs law
]tex]F = \frac{kq_1q_2}{r^2}[/tex]
here we have




now by using the above equation we have


so here the force between two charges is of above magnitude and this will be repulsive force between them as both charges are of same sign.
Question
What is the length of the pipe?
Answer:
(a) 0.52m
(b) f2=640 Hz and f3=960 Hz
(c) 352.9 Hz
Explanation:
For an open pipe, the velocity is given by

Making L the subject then

Where f is the frequency, L is the length, n is harmonic number, v is velocity
Substituting 1 for n, 320 Hz for f and 331 m/s for v then

(b)
The next two harmonics is given by
f2=2fi
f3=3fi
f2=3*320=640 Hz
f3=3*320=960 Hz
Alternatively,
and 

(c)
When v=367 m/s then
