Answer:
[tex]Q = -1.43\times 10^[-5} coulomb[/tex]
Explanation:
Given data:
particle mass = 0.923 g
particle charge is 4.52 micro C
speed of particle 45.7 m/s
In this particular case, coulomb attraction will cause centrifugal force and taken as +ve and Q is taken as -ve
[tex]-\frac{Qq}{4\pi \epsilon r^2} = \frac{mv^2}{r}[/tex]
solving for Q WE GET
[tex]Q = -\frac{mv^2}{r} \times r^2 \frac{4\pi \epsilon}{q}[/tex]
[tex]Q = -mv^2\times r \frac{4\pi \epsilon}{q}[/tex]
[tex]Q = - \frac{0.923\times 10^{-3} \times 45.7^2\times (22.6\times 10^{-2})} {4.52\times 10^{-6} \times 9\times 10^9}[/tex]
where[tex] \frac{1}{4\pi \epsilon} = 9\times 10^9[/tex]
[tex]Q = -1.43\times 10^[-5} coulomb[/tex]
