Respuesta :
Answer:
The magnitude of the charge on each sphere is 0.135 μC
Explanation:
Given that,
Mass = 1.0
Distance = 2.0 cm
Acceleration = 414 m/s²
We need to calculate the magnitude of charge
Using newton's second law
[tex]F= ma[/tex]
[tex]a=\dfrac{F}{m}[/tex]
Put the value of F
[tex]a=\dfrac{kq^2}{mr^2}[/tex]
Put the value into the formula
[tex]414=\dfrac{9\times10^{9}\times q^2}{1.0\times10^{-3}\times(2.0\times10^{-2})^2}[/tex]
[tex]q^2=\dfrac{414\times1.0\times10^{-3}\times(2.0\times10^{-2})^2}{9\times10^{9}}[/tex]
[tex]q^2=1.84\times10^{-14}[/tex]
[tex]q=0.135\times10^{-6}\ C[/tex]
[tex]q=0.135\ \mu C[/tex]
Hence, The magnitude of the charge on each sphere is 0.135μC.
Answer:
Charge, [tex]q=1.35\times 10^{-7}\ C[/tex]
Explanation:
It is given that,
Mass of spheres, m = 1 g = 0.001 kg
Distance between spheres, r = 2 cm = 0.02 m
Acceleration when they released, [tex]a=414\ m/s^2[/tex]
We need to find the magnitude of the charge on each sphere. The electric force is given by :
[tex]F=\dfrac{kq^2}{r^2}[/tex]
Also, F = ma
[tex]ma=\dfrac{kq^2}{r^2}[/tex]
[tex]q^2=\dfrac{mar^2}{k}[/tex]
[tex]q^2=\dfrac{0.001\times 414\times (0.02)^2}{9\times 10^9}[/tex]
[tex]q=1.35\times 10^{-7}\ C[/tex]
So, the magnitude of charge on each spheres is [tex]1.35\times 10^{-7}\ C[/tex]. Hence, this is the required solution.
