AP EAMCET2003PhysicsElectrostatics
An infinite number of electric charges each equal to 5 nano-coulomb (magnitude) are placed along X -axis at x=1 ~cm , x=2 ~cm , x=4 ~cm , x=8 ~cm . . . . and so on. In this set up if the consecutive charges have opposite sign, then the electric field in newton/coulomb at x=0 is : ( 1 4 ₀ =9 10^9 ~N - m ^2 / C ^2 )
Options
- A12 10^4
- B24 10^4
- C36 10^4
- D48 10^4
Correct answer
C. 36 10^4
Step-by-step solution
Electric field intensity due to a point charge. E= 1 4 ₀ Q r^2 The consecutive charges are of opposite signs. Net electric field at x=0 , is E= 1 4 ₀ [ Q r₁^2 - Q r₂^2 + Q r₃^2 - Q r₄^2 . . ]= Q 4 ₀ [ 1 r₁^2 - 1 r₂^2 + 1 r₃^2 - 1 r₄^2 + . =9 10^9 5 10⁻⁹ [ 1 (1 10⁻² )^2 - 1 (2 10⁻² )^2 + 1 (4 10⁻² )^2 + = 45 10⁻⁴ [ 1 1^2 - 1 2^2 + 1 4^2 - 1 8^2 + . ]=45 10^4 [ 1 1^2 - 1 2^2 + 1 4^2 - 1 8^2 ] 1 1^2 - 1 2^2 + 1 4^2 - 1 8^2 = 1 1^2 1- ( -1 2 )^2 [ . Sum of infinite GP, . S _ = a 1-r ]= 1 1+ 1 4 = 4 5 E=45 10^4 4 5 =36