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# Solve it, - Electrostatics - JEE Main-4

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)

0.9 N

• Option 2)

1.8 N

• Option 3)

2.7 N

• Option 4)

3.6 N

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As we learned

Magnitude of the Resultant force -

$\dpi{100} F_{net}=\sqrt{F_{1}^{2}+F_{2}^{2}+2F_{1}F_{2}\cos \Theta }$

- wherein

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

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

$F_{B}$ = 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)

0.9 N

Option 2)

1.8 N

Option 3)

2.7 N

Option 4)

3.6 N

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