Question

What is the force between a 0.004 C charge and a -0.02 C charge separated by a distance of 6 m

Answer 1

We can solve this problem applying "Coulomb's Law" which states-

`\:\:\:\:\:\:\:\:\:\:\star\:\sf \underline{F_((air)) = K* ( q_1 q_2)/(r^2)} \\`

`\:\:\:\:\:\:\:\:\:\:\star\:\sf\underline{F_((air)) = (1)/(4\pi \epsilon_(0) ) (q_1q_2)/(r^2) }\\`

Where-

• q₁ and q₂ are the two cahrges.
• r is the distance between the charges.
• 
  `\sf \epsilon_(0)` is the permittivity of free space.
• K is the Coulomb's Constant.
• k = 9×10⁹ Nm²/C²

As per question, we are given that -

• q₁= 0.004C
• q₂= -0.02 C
• Distance,r = 6 m

Now that required values are given, so we can substitute the values into the formula and solve for Force -

`\:\:\:\:\:\:\:\:\:\:\star\:\sf\underline{F_((air)) = (1)/(4\pi \epsilon_(0) ) (q_1q_2)/(r^2) }\\`

`\:\:\:\:\:\:\:\longrightarrow \sf Force_((air))= 9* 10^9 * (0.004* -0.02 )/((6)^2)\\`

`\:\:\:\:\:\:\:\longrightarrow \sf Force_((air))= -9* 10^9 *(0.00008)/(36)\\`

`\:\:\:\:\:\:\:\longrightarrow \sf Force_((air)) = -9* 10^9 * 2.2* 10^(-6)\\`

`\:\:\:\:\:\:\:\longrightarrow \sf \underline{Force_((air)) = -19800\:N}\\`

• Therefore, the force between two charges is -19800 N.