three charges + 2 uC, + 3 uC and +4uC are placed at the corners of the equilateral triangle ABC for each of side 0.2 M find the force on +4uC
due to other two charges
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Answered by
32
The charges will have to be at a specific distance and hence would react . If u did not got it ask. This is just a hint.
K = 1/(4πε) = 9 * 10⁹ N-m²/C²
forces F1 and F2 on the charge at Q are along the sides AB and AC.
magnitude of F1 = F2 = K * 2 * 10⁻⁶ * 3 * 10⁻⁶ / 0.20²
= 1.5 * 10⁻⁸ K Newtons
components along the horizontal directions are equal to F1 Cos 60° and F2 Cos 60° which are equal in magnitude and opposite in direction. They cancel.
Their components along AM will add to :
2 F1 Cos 30° = 2 * 1.5 * 10⁻⁸ * √3/2 * K newtons = 1.5√3 *10⁻⁸ K Newtons
Let the charge placed at M be q Coulombs. AM² = 0.20² - 0.10² = 0.03 m²
Force due to q on charge at A will be along MA: F3
F3 = K q * 2 * 10⁻⁶ / 0.03 = 6.667 q K * 10⁻⁹ Newtons
For the charge at A to be stationary,
6.667 q K * 10⁻⁹ = 1.5 √3 * 10⁻⁸ K
q = 3.897 Coulombs positive charge.
K = 1/(4πε) = 9 * 10⁹ N-m²/C²
forces F1 and F2 on the charge at Q are along the sides AB and AC.
magnitude of F1 = F2 = K * 2 * 10⁻⁶ * 3 * 10⁻⁶ / 0.20²
= 1.5 * 10⁻⁸ K Newtons
components along the horizontal directions are equal to F1 Cos 60° and F2 Cos 60° which are equal in magnitude and opposite in direction. They cancel.
Their components along AM will add to :
2 F1 Cos 30° = 2 * 1.5 * 10⁻⁸ * √3/2 * K newtons = 1.5√3 *10⁻⁸ K Newtons
Let the charge placed at M be q Coulombs. AM² = 0.20² - 0.10² = 0.03 m²
Force due to q on charge at A will be along MA: F3
F3 = K q * 2 * 10⁻⁶ / 0.03 = 6.667 q K * 10⁻⁹ Newtons
For the charge at A to be stationary,
6.667 q K * 10⁻⁹ = 1.5 √3 * 10⁻⁸ K
q = 3.897 Coulombs positive charge.
Answered by
8
Answer: the net force on +4uC charge is 3.923 N
Explanation:
Attachments:
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