a point charge Q is situated at point B on the ground. A point charge Q of mass m is vertically dropped along line Ab from a multi storey building of height h. find its position when it is in equilibrium
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We are assuming negligible air resistance.
Final Answer : (4π£°mg/Q^2)^1/2 from Point B.
Steps and Calculation :
1) Initial the charge will fall vertically downward due to excess in gravitational force.
2) After some time, it's velocity reaches zero as Electrostatic force increases when charges came close.
At that instant, These two forces are in equilibrium.
2 )Let that position be at (x) units from point B.
Here, F net = 0
=> F electrostatic - F gravity =0
=> KQ^2 / x^2. = mg
=> x = (4π£°mg) ^1/2 / Q
Final Answer : (4π£°mg/Q^2)^1/2 from Point B.
Steps and Calculation :
1) Initial the charge will fall vertically downward due to excess in gravitational force.
2) After some time, it's velocity reaches zero as Electrostatic force increases when charges came close.
At that instant, These two forces are in equilibrium.
2 )Let that position be at (x) units from point B.
Here, F net = 0
=> F electrostatic - F gravity =0
=> KQ^2 / x^2. = mg
=> x = (4π£°mg) ^1/2 / Q
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