Question

Three charges, each separated by 100 m from adjacent charges, are located along a horizontal line: a -3.00 C charge on the left, a 2.00 C charge in the middle, and a 1.00 C charge on the right. What is the resultant force on the 1.00 C charge due to the other two

Answer 1

Answer:

F= 11.25*10⁵ N to the right.

Explanation:

• Assuming that the three charges can be treated like point charges, they must obey Coulomb's Law.
• Due to the linearity of this Law, we can use superposition in order to find the resultant force on the 1.00 C charge due to the other two.
• First, we find the force that the -3.00 C charge (located 200 m to the left) exerts on the 1.00 C, as follows:

      $F_{13} = \frac{K*q_{1}*q_{2}}{r_{13} ^{2} } = \frac{9e9*(-3.00C)(1.00C)}{(200m)^{2}} = -6.75e5 N (1)$

• Then, in the same way, we can find the force that the 2.00 C exerts on the 1.00 C charge, located 100 m away to the left:

      $F_{23} = \frac{K*q_{3}*q_{2}}{r_{23} ^{2} } = \frac{9e9*(2.00C)(1.00C)}{(100m)^{2}} = 18e5 N (2)$

• Since both vectors are on the same line, their sum is directly the algebraic sum, as follows:
• F₃ = F₁₃ + F₂₃ = -6.75*10⁵ N + 18.00*10⁵ N = 11.25*10⁵ N to the right, assuming this direction as positive.