Question

Three positive charges 3nC ,4nC and 5nC are placed at the vertices A,B and C respectively of an equilateral triangle ABC of side of 0.2m.find only the magnitude of the force on the largest charge

Answer 1

Answer:so force experienced by A due to C is 33.75 × 10-7N and force experienced by B due to C is 45×10-7N

Explanation:

Answer 2

Answer:

The magnitude of force on the largest charge is 6.84 x 10⁻⁶ N .

Explanation:

Given that :

• 3nC, 4nC, 5nC charges are placed at vertices of equilateral triangle ABC.
• length of each side of triangle = 0.2 m

To find :

• Force on 5nC charge due to other charges

Solution :

• Let, the force experience by charge 5nC due to charge 3 nC placed at point A be F1.

$F1 = \frac{1}{4\piε₀} \frac{5 \times {10}^{ - 9} \times 3 \times {10}^{ - 9} }{ {0.2}^{2} } \\ = \frac{1}{4\piε₀} \frac{5 \times 3 \times {10}^{- 18} }{4 \times {10}^{ - 2} }$

• Now, let the force experience by charge 5nC due to charge 4 nC placed at point B be F2.

$F2= \frac{1}{4\piε₀} \frac{5 \times {10}^{ - 9} \times 4\times {10}^{ - 9} }{ {0.2}^{2} } \\ = \frac{1}{4\piε₀} \frac{5 \times 4 \times {10}^{- 18} }{4 \times {10}^{ - 2} }$

• Let, for simplicity of solving, we assume

$k = \frac{1}{4\piε₀} \frac{5 \times {10}^{ - 18} }{ {4 \times 10}^{ - 2} } \\ = 9 \times {10}^{9} \times 1.25 \times {10}^{ - 16} \\ = 1.125 \times {10}^{ - 6}$

• The force F1 becomes, F1 = 3k.
• The force F2 becomes, F2 = 4k.
• The net force experience by charge 5nC placed at point C due to charge 3nC and 4nC is vector sum of force F1 and F2.
• The angle between F1 and F2 is 60⁰.
• The net resultant force, F is

$F = \sqrt{ {3k}^{2} + {4k}^{2} + 2(3k)(4k) \cos60⁰ } \\ = k\sqrt{9 + 16 + 12} \\ = k\sqrt{37} \\ = 6.08k$

• Putting value of k in above equation, we get;
• F = 6.08 x 1.125 x 10⁻⁶ N = 6.84 x 10⁻⁶ N .