a charge of 1.6 × 10^-19 C is situated at a distance of 30 cm from another charge of 6.4 × 10^-19C where will be the electric field be zero?
Answers
Given:
Repulsive force of magnitude 6 × 10−3 N
Charge on the first sphere, q1 = 2 × 10−7 C
Charge on the second sphere, q2 = 3 × 10−7 C
Distance between the spheres, r = 30 cm = 0.3 m
Electrostatic force between the spheres is given by the relation:
F = (1/4πε0). (q1q2)/ (r2)
Where, ε0 = Permittivity of free space and (1/4πε0) =9 × 109 Nm2C−2
Therefore, force F = (9 × 109 × 2 × 10−7)/ ((0.3)2)
= 6 × 10−3N
Solution:- Let P be the point where the electric field is zero. Let P be "x" metre from the charge 1.6 × 10^-19 C. Then a charge 6.4 × 10^-19 C is (0.3 - x) m from P. Electric field at point P due to charge of 1.6 × 10^-19 C
A._____.P________________.B
<=x m =><= (0.3 - x) m=>
<= 0.3 metre =>
= 9×10^9× 1.6 × 10^-19 C/x^2 N/C
towards B
and electric field at P due to charge 6.4 × 10^-19C is
= 9×10^9× 1.6 × 10^-19 C/ (0.3 - x) N/C
towards A
For field to be zero at P,
=> 9×10^9× 1.6 × 10^-19 C/x^2 N/C = 9×10^9× 1.6 × 10^-19 C/ (0.3 - x) N/C
=> (0.3 - x)^2 = 4x^2
=> 0.3 - x = 2x
=> 3x = 0.3
=> x = 0.1 m
Thus, the electric field is zero between the two charges, and 0.1 m from the charge 1.6 × 10^-19 C.