KCET2017PhysicsWave Optics
Two point charges ( A=+3 nC ) and ( B=+1 nC ) are placed ( 5 ~cm ) apart in air. The work done to move charge ( B ) towards ( A ) by ( 1 ~cm ) is
Options
- A( 1.35 10⁻⁷ ~J )
- B( 2.7 10⁻⁷ ~J )
- C( 2.0 10⁻⁷ ~J )
- D( 12.1 10⁻⁷ ~J )
Correct answer
A. ( 1.35 10⁻⁷ ~J )
Step-by-step solution
Given A=3 n C=3 10⁻⁹ C ; B=1 n C=1 10⁻⁹ C ; distance, d=5 ~cm =5 10⁻² ~m Therefore, energy of system U_ i = 1 4 ₀ q_ A q_ B d = (9 10⁹ ) (3 10⁻⁹ ) (1 10⁻⁹ ) (5 10⁻² ) To move charge B towards A by 1 ~cm d^ =4 ~cm =4 10⁻² ~m Therefore, energy of system after moving charge is U_ f = 1 4 ₀ q_ A q_ B d = (9 10⁹ ) (3 10⁻⁹ ) (1 10⁻⁹ ) (4 10⁻² ) Therefore, work done =U_ f -U_ i = (9 10⁹ ) (3 10⁻⁹ ) (1 10⁻⁹ ) (4 10⁻¹² ) - (9 10⁹ ) (3 10⁻⁹ ) (1 10⁻⁹ ) (5 10⁻² ) = (9 10⁹ ) (3 10⁻⁹ ) (1 10⁻⁹ ) [ 1 4 10⁻² - 1 5 10⁻² ]= 27 10⁻⁹