The magnitude and direction of the electrostatic torce is - 0.00712 x 10^9N towards left direction
Given:
q1 = +2 uC
q2 = -2 uC
q3 = +4 uC
To Find:
magnitude le and direction of the electrostatic torce
Solution: Electric force is a vector quantity. The electric force is proportional to the product of the magnitude of the charges. It is inversely proportional to the square of the distance between the charges. The electric force is calculated by,
F = kqQ/r^2
Force between q1 and q2
F1 = +2 x -2 x k/50 x 50 = - 0.0016k
Force between q1 and q3
F2 = +2 x +4 x k/100 x 100 = + 0.0008k
Net force on charge at origin is F1 + F2
F(net) = F1 + F2 = - 0.0016k + 0.0008k
F(net) = - 0.0008 x 8.9 x 10^9 = - 0.00712 x 10^9N
So, force on charge at origin is - 0.00712 x 10^9N and towards left direction
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Answer:
the period T of whole motion should be twice the value for half at he bottom so T is 0.2sec.
w is angular frequency
formula:2π/T
now k is spring constant
F/R-->mw²
putting values:70*(2π/0.2)²
=4.9x10⁶
so we can say that SHM is not affected by the amplitude of the bounce.
