Question

2 point charges 3nc and -3nc are placed at two cornners A and B of an eqilateral triangle of side 0.1 m Find the resultant electric field ​

Answer 1

Answer:

8.1×10^(-6) Newton the force of attraction

Explanation:

q1 = 3nc. (3 nano c)

q2= -3nc ( -3nano c)

r= 0.1 m ( corner A and B are the end pts of AB)

F = (1/4π€0) q1q2/r²

= 9×10^9× 3 ×10^(-9)×(-3)× 10^(-9) / (0.1)²

= 9×9×10^( 9-9-9+2)

= 81×10^(-7)

= -8.1×10^(-6) Newton

hope this helps you