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# Need explanation for: - Electrostatics - JEE Main-5

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

• Option 1)

$0.9N\; \;$

• Option 2)

$\; 1.8N\; \;$

• Option 3)

$\; 2.7N\;$

• Option 4)

$\; 3.6N$

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As we learned

By conduction -

When two conducters brought in contact.

- wherein

i.e The charge will spread over both the conducters.

$F_{A}=force \; on \; C\; due\; to\; charge \; placed\; at\; A$

$=9\times 10^{9}\times \frac{10^{-6}\times2\times10^{-6}}{(10\times10^{-2})^{2} }=1.8N$

FB = force on C due to charge placed at B

$=9\times 10^{9}\times \frac{10^{-6}\times2\times10^{-6}}{(0.1)^{2} }=1.8N$

Net force on C

$F_{net}=\sqrt{(F_{A})^{2}+(F_{B})^{2}+2F_{A}F_{B}\cos 120^{\circ}}=1.8N$

Option 1)

$0.9N\; \;$

Option 2)

$\; 1.8N\; \;$

Option 3)

$\; 2.7N\;$

Option 4)

$\; 3.6N$

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