Question

Two point Charges q1
and q2
separated by a distance of 3.0 m experience a
mutual force of 16 × 10–15 N. Calculate the magnitude of force when q1
= q2
= q. What will be the magnitude of force if separation distance is changed to
6.0 m?

Answer 1

Two point Charges q1 and q2 separated by a distance of 3.0 m experience a

mutual force of 16 × 10–15 N. Calculate the magnitude of force when q1= q2 = q. What will be the magnitude of force if separation distance is changed to 6.0 m?

solution :

using Coulomb's force,

F = kq1.q2/r²

if q1 = q2 = q

F = kq²/r²

given, F = 16 × 10^-15 N , r = 3m

so, 16 × 10^-15 = (9 × 10^9 × q²)/(3)²

⇒16 × 10^-15 = 10^9 × q²

⇒16 × 10^-24 = q² .........(1)

now magnitude of force when seperation between charges is 6m

so, F' = kq²/r'²

= (9 × 10^9 × 16 × 10^-24)/(6)²

= 4 × 10^-15 N

hence, magnitude of force is 4 × 10^-15 N

Answer 2

$\huge\bold\purple{Answer:-}$

using Coulomb's force,

F = kq1.q2/r²

if q1 = q2 = q

F = kq²/r²

given, F = 16 × 10^-15 N , r = 3m

so, 16 × 10^-15 = (9 × 10^9 × q²)/(3)²

⇒16 × 10^-15 = 10^9 × q²

⇒16 × 10^-24 = q² .........(1)

now magnitude of force when seperation between charges is 6m

so, F' = kq²/r'²

= (9 × 10^9 × 16 × 10^-24)/(6)²

= 4 × 10^-15 N

hence, magnitude of force is 4 × 10^-15 N