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

${x}^{2} + y \leqslant 2$

this is the answer

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.