Respuesta :
Answer:
The charge on point A is [tex]q_a[/tex] = 2.4 × [tex]10^{-5}[/tex] C
The charge on point B is [tex]q_b[/tex] = 1.2 × [tex]10^{-5}[/tex] C
Explanation:
Given data
Distance (r) = 0.19 m
Magnitude of the charge on A is twice that of the charge on B i.e.
[tex]q_A = 2 q_B[/tex]
F = 46 N
We know that force between the charges is given by
[tex]F = \frac{k Q_A Q_B}{r^{2} }[/tex]
[tex]65 = \frac{(9) (10^{9}) (2q_b^{2} ) )}{0.2^{2} }[/tex]
[tex]q_b[/tex] = 1.2 × [tex]10^{-5}[/tex] C
[tex]q_a[/tex] = 2 × [tex]q_b[/tex]
[tex]q_a[/tex] = 2 × 1.2 × [tex]10^{-5}[/tex]
[tex]q_a[/tex] = 2.4 × [tex]10^{-5}[/tex] C
Therefore the charge on point A is [tex]q_a[/tex] = 2.4 × [tex]10^{-5}[/tex] C
The charge on point B is [tex]q_b[/tex] = 1.2 × [tex]10^{-5}[/tex] C
