Respuesta :
Answer:
2.92 x 10¹² electrons
Explanation:
given,
distance between two plastic ball, r = 17 cm
r = 0.17 m
Force of attraction = F = 68 mN
F = 0.068 N
number of electron transferred from one ball to another.
using Coulomb Force equation
[tex]F = \dfrac{kq^2}{r^2}[/tex]
[tex]0.068 = \dfrac{9\times 10^9\times q^2}{0.17^2}[/tex]
q² = 2.1835 x 10⁻¹³
q = 4.67 x 10⁻⁷ C
now, number of electron
[tex]N = \dfrac{q}{e}[/tex]
e is the charge of electron which is equal to 1.6 x 10⁻¹⁹ C
[tex]N = \dfrac{4.67\times 10^{-7}}{1.6\times 10^{-19}}[/tex]
N = 2.92 x 10¹² electrons
electrons were transferred from one ball to the other is 2.92 x 10¹²
Field strength is directly proportional to the square of charge. The charge on the given plastic ball is [tex]4.67 \times 10^{-7} \rm \ C[/tex]
From Coulomb law:
[tex]F = \dfrac {kQ^2}{r^2}[/tex]
Where,
[tex]F[/tex]- field strength = 68 mN = 0.06 N
[tex]Q[/tex]- charge
[tex]r[/tex]- distance = 17 cm =0.17 m
[tex]k[/tex]- constant = 9x10⁹
Put the values in the formula,
[tex]0.068 = \dfrac{9\times 10^9 \rimes Q^2}{0.017^2}\\\\q^2 = 2.1835 \times 10^{-13}\\\\q = 4.67 \times 10^{-7} \rm \ C[/tex]
Therefore, the charge on the given plastic ball is [tex]4.67 \times 10^{-7} \rm \ C[/tex].
To know more about Coulomb law:
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