Three charges (–Q, +2Q, –2Q) are placed at the vertices of an equilateral triangle of side length, [math: 1.0 × 10−6 m], as shown below. Q represents [math: 1.0 × 10−10 C].What is the magnitude and direction of the net electrostatic force exerted on charge –Q by both charges +2Q and -2Q? Give your answer in standard form to 2 decimal places without units followed by the direction.Example format: [math: 1.23 × 1045 N Right] must be written as [ 1.23e45 Right ][math: 4.32 × 10−1 N Up] must be written as [ 4.32e-1 Up ]

Question

Three charges (–Q, +2Q, –2Q) are placed at the vertices of an equilateral triangle of side length, [math: 1.0 × 10−6 m], as shown below. Q represents [math: 1.0 × 10−10 C].What is the magnitude and direction of the net electrostatic force exerted on charge –Q by both charges +2Q and -2Q? Give your answer in standard form to 2 decimal places without units followed by the direction.Example format: [math: 1.23 × 1045 N Right] must be written as [ 1.23e45 Right ][math: 4.32 × 10−1 N Up] must be written as [ 4.32e-1 Up ]

BlackTom AI · Accepted Answer

Coulomb’s law gives the magnitude of the force between two point charges: F = k|q1 q2|/r^2. Use vector components to sum the two forces from the lower charges; symmetry of the equilateral triangle simplifies the result.

1. Relevant concepts
- Coulomb’s law (k = 8.99×10^9 N·m^2/C^2).
- Vector addition and components. In an equilateral triangle of side a, the vertical height is h = (√3/2)a and the horizontal projections of the side are a/2.
- Attraction/repulsion directions: +2Q attracts -Q; -2Q repels -Q.

2. Step-by-step reasoning and intermediate steps
- Given: a = 1.0×10−6 m, Q = 1.0×10−10 C.
- Magnitude of each pairwise force (top with either bottom charge):
  F0 = k·(2Q·Q)/a^2 = k·2Q^2 / a^2.
  Substitute numbers: Q^2 = (1.0×10−10)^2 = 1.0×10−20; a^2 = (1.0×10−6)^2 = 1.0×10−12.
  So F0 = (8.99×10^9)·2·(1.0×10−20)/(1.0×10−12) = 2·8.99×10^9·1.0×10−8 = 2·8.99×10^1 = 179.8 N.
- Components: each force has horizontal component F0·(1/2) and vertical component F0·(√3/2). The left-bottom +2Q attracts the top charge giving a leftward and downward component (x negative, y negative). The right-bottom −2Q repels the top charge giving a leftward and upward component (x negative, y positive).
- Vertical components cancel: −F0·(√3/2) + F0·(√3/2) = 0.
- Horizontal components add: −F0·(1/2) + −F0·(1/2) = −F0. Net force magnitude = F0 = 179.8 N, direction = left.

3. Rounding and final formatting
- 179.8 N = 1.798×10^2 N. To two decimal places in standard form: 1.80e2.
- Other possible directions (Right, Up, Down) are incorrect because the horizontal components point left and vertical components cancel by symmetry; any other magnitude would come from arithmetic or geometry errors.

Conclusion: Net force = 1.80e2 Left, which matches the provided answer.

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