Calculate the force between 3×10^-6C and 4×10^-8C at 4cm. Take 1÷4piE0 = 9×10^9

1 answer

To calculate the force between two point charges, we can use Coulomb's Law, which is given by the formula:

\[ F = k \frac{|q_1 q_2|}{r^2} \]

where:

  • \( F \) is the force between the charges,
  • \( k = \frac{1}{4\pi \epsilon_0} \) is the Coulomb's constant (given as \( 9 \times 10^9 , \text{N m}^2/\text{C}^2 \)),
  • \( q_1 \) and \( q_2 \) are the charges,
  • \( r \) is the distance between the charges.

Given:

  • \( q_1 = 3 \times 10^{-6} , \text{C} \)
  • \( q_2 = 4 \times 10^{-8} , \text{C} \)
  • \( r = 4 , \text{cm} = 0.04 , \text{m} \)

Now, we can substitute these values into Coulomb's Law:

  1. Calculate the product of the charges: \[ |q_1 q_2| = |(3 \times 10^{-6}) (4 \times 10^{-8})| = 12 \times 10^{-14} , \text{C}^2 \]

  2. Calculate \( r^2 \): \[ r^2 = (0.04)^2 = 0.0016 , \text{m}^2 \]

  3. Now plug in the values into Coulomb's Law: \[ F = 9 \times 10^9 \frac{12 \times 10^{-14}}{0.0016} \]

  4. Calculate the fraction: \[ \frac{12 \times 10^{-14}}{0.0016} = \frac{12 \times 10^{-14}}{1.6 \times 10^{-3}} = 7.5 \times 10^{-11} \]

  5. Now calculate \( F \): \[ F = 9 \times 10^9 \times 7.5 \times 10^{-11} = 6.75 \times 10^{-1} , \text{N} = 0.675 , \text{N} \]

Thus, the force between the two charges is 0.675 N.