Electrostatics Question 195

Question: Electric charges of $ 1\mu C,-1\mu C $ and $ 2\mu C $ are placed in air at the corners A, B and C respectively of an equilateral triangle ABC having length of each side 10 cm. The resultant force on the charge at C is [EAMCET (Engg.) 2000]

Options:

A) 0.9 N

B) 1.8 N

C) 2.7 N

D) 3.6 N

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Answer:

Correct Answer: B

Solution:

FA = force on C due to charge placed at A

$ =9\times 10^{9}\times \frac{{{10}^{-6}}\times 2\times {{10}^{-6}}}{{{(10\times {{10}^{-2}})}^{2}}}=1.8N $

FB = force on C due to charge placed at B

$ =9\times 10^{9}\times \frac{{{10}^{-6}}\times 2\times {{10}^{-6}}}{{{(0.1)}^{2}}}=1.8N $

Net force on C $ F _{net}=\sqrt{{{(F _{A})}^{2}}+{{(F _{B})}^{2}}+2F _{A}F _{B}\cos 120^{o}}=1.8N $



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