Question

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?​

Answer 1

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

Answer 2

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.