Question

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​

Answer 1

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.

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