Three charges are on a line. A positive charge is on the far left labeled q Subscript 1 baseline positive 6 Coulombs. The second charge is 2 m away and is labeled q Subscript 2 baseline negative 4 Coulombs. The third charge is 1 m away and is labeled q Subscript 3 baseline = positive 3 Coulombs.

What is the electrical force between q1 and q2? Recall that k = 8.99 × 109 N•meters squared over Coulombs squared..

4.3 × 1010 N
3.5 × 1010 N
–5.4 × 1010 N
–5.8 × 1010 N

1 answer

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

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

where:

  • \( F \) is the magnitude of the force between the charges,
  • \( k \) is Coulomb's constant (\( 8.99 \times 10^9 , \text{N m}^2/\text{C}^2 \)),
  • \( q_1 \) and \( q_2 \) are the magnitudes of the charges,
  • \( r \) is the distance between the charges.

Given:

  • \( q_1 = +6 , \text{C} \)
  • \( q_2 = -4 , \text{C} \)
  • The distance \( r = 2 , \text{m} \)

Plugging the values into the formula:

\[ F = (8.99 \times 10^9) \frac{|(6)(-4)|}{(2)^2} \]

Calculating the values step by step:

  1. Calculate the product of the charges: \[ |q_1 q_2| = |6 \times -4| = 24 , \text{C}^2 \]

  2. Calculate \( r^2 \): \[ r^2 = 2^2 = 4 , \text{m}^2 \]

  3. Now substitute these values back into the force equation: \[ F = (8.99 \times 10^9) \frac{24}{4} \]

    Simplifying inside the fraction: \[ \frac{24}{4} = 6 \]

  4. Now calculate \( F \): \[ F = (8.99 \times 10^9) \times 6 \] \[ F = 53.94 \times 10^9 , \text{N} \] \[ F = 5.394 \times 10^{10} , \text{N} \]

Since \( q_1 \) is positive and \( q_2 \) is negative, the force is attractive, which means that we should assign a negative sign to the force:

\[ F = -5.394 \times 10^{10} , \text{N} \]

Thus, rounding to the appropriate number of significant figures, we can express the final answer as approximately:

\[ F \approx -5.4 \times 10^{10} , \text{N} \]

Therefore, the correct answer is:

–5.4 × 10^10 N