1. Two point charges of +10uc and -10pc are placed at a
distance 40 cm in air. Potential energy of the system will
be:
[BSEB, 2019 (A)]
(A) 2.25 )
(B) 2.35 J
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Given info : Two point charges of +10μC and -10μC are placed at a distance of 40cm in air.
To find : Potential energy of the system will be...
solution : potential energy of the system is given by, U = kq₁q₂/r
here q₁ = +10μC = 10¯⁵C , q₂ = -10μC = -10¯⁵C
r = 40cm = 0.4 m
so, U = (9 × 10^9 × 10¯⁵ × -10¯⁵)/(0.4)
= -(9 × 10¯¹)/(0.4)
= -9/4 = -2.25 J
Therefore the potential energy of the system is -2.25 J.
also read similar questions : Two charges 10uC and 5uC are placed 1m apart. In order to make this distance 0.5cm, how much work would be done?
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Force acts on charge 5uC due to charge 10uC is F1, and force acts on charge 10uC due to charge 5uC is F2 for same separa...
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