Question

Hey Guys Help Please♥️
Charges q1 = 100 u C and q2= 50 u C are located in xy-plane at positions r1 = 3.0 j and
r2 = 4.0 i respectively. Where the distance is measured in meters. Calculate the force
on q2
(a) 1.8 N
(b) 2.5 N
(C) ZN
(d) 8N
​

Answer 1

Answer:

9N

Explanation:

distance between the charges q1 and q2

__________

= √(0-4)²+(3-0)²

= √25

r = 5 meters

given q1 = 100uC =1× 10^(-4) C

q2 = 50uC = 5×10^(-5) C

=> Force(F) = (1/4π€0) q1q2/r²

= 9×10^9×10^(-4)×5×10(-5)/5

= 9 N

=> all options are false.

=> The force between the charges is 9N

hope this helps you