Question

Two charges \(+4\,\mu C\) and \(+9\,\mu C\) are separated by a distance of \(3\,m\) in vacuum. The electrostatic force between them is \((k=9\times10^9\,Nm^2C^{-2})\)

• A. \(0.018\,N\)
• B. \(0.036\,N\)
• C. \(0.054\,N\)
• D. \(0.072\,N\)

Hint:

Coulomb's Law: \[ \boxed{F=\frac{kq_1q_2}{r^2}} \] Important: \[ F\propto q_1q_2 \] \[ F\propto \frac{1}{r^2} \] Doubling distance reduces force by a factor of 4.

Answer 1

The Correct Option is B

Step 1: Recall Coulomb's law. \[ F=\frac{kq_1q_2}{r^2} \] Given: \[ q_1=4\times10^{-6}C \] \[ q_2=9\times10^{-6}C \] \[ r=3m \]

Step 2: Substitute the values. \[ F= \frac{(9\times10^9)(4\times10^{-6})(9\times10^{-6})} {3^2} \] \[ F= \frac{9\times36\times10^{-3}} {9} \] \[ F=36\times10^{-3} \] \[ F=0.036N \]

Step 3: Choose the correct option. \[ \boxed{F=0.036N} \] Therefore, \[ \boxed{\text{(B)}} \] is the correct answer.