Question

1. Two point charges, q1 and q2, are located 5 cm apart. The magnitude of q1 is 3 μC and the magnitude of q2 is -5 μC. What is the force between these charges, according to Coulomb's law?

Answer 1

The force between these charges, according to Coulomb's law would be -54 N.

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

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

`\qquad\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

According to the given parameters -

• Magnitude of q₁= 3 μC
• Magnitude of q₂= -5 μC
• Distance,r = 5cm

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

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

`\longrightarrow \sf Force = 9* 10^9 * ( 3* 10^(-6)* -5 * 10^(-6))/(\bigg(5* 10^(-2)\bigg)^2)\\`

`\longrightarrow \sf Force = -(9* 5* 3* 10^(9) * 10^(-12))/(25* 10^(-4))\\`

`\qquad\longrightarrow \sf Force =- (135 * 10^(9-12+4))/(25)\\`

`\qquad\longrightarrow \sf Force = - \frac{\cancel{135}}{\cancel{25}}* 10\\`

`\qquad\longrightarrow \sf Force = -5.4 * 10\\`

`\qquad\qquad\longrightarrow \sf \underline{Force = \boxed{\sf{-54 N}}} \\`

• Henceforth, The force between these charges, according to Coulomb's law would be -54 N.