Three equal charges, each having a magnitude of $4 μ C$ , are placed at the three corners of a right-angled triangle of sides $6 c m, 8 c m$ and $10 c m .$ The force on the charge at the right-angle corner will be

11.5 N

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23 N

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46 N

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230 N

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Solution

The correct option is A $11.5 N$
Two forces are,

$F_{1} = \frac{k Q_{1} Q_{2}}{{R_{1}}^{2}}$

$= \frac{9 \times 10^{9} \times {(4 \times 10^{- 6})}^{2}}{{(6 \times 10^{- 2})}^{2}}$

$= 10 N$

$F_{2} = \frac{k Q_{1} Q_{2}}{{R_{2}}^{2}}$

$= \frac{9 \times 10^{9} \times {(4 \times 10^{- 6})}^{2}}{{(8 \times 10^{- 2})}^{2}}$

$= \frac{360}{64} = 5.62 N$

Resultant force at the right angle vertex is,

$F = \sqrt{({F_{1}}^{2} + {F_{2}}^{2})}$

$= \sqrt{10^{2} + {5.62}^{2}}$

$= \sqrt{131.56}$

$= 11.5 N$

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Three equal charges, each having a magnitude of 4μC , are placed at the three corners of a right-angled triangle of sides 6cm,8cm and 10cm. The force on the charge at the right-angle corner will be

The correct option is A 11.5NTwo forces are, F1=kQ1Q2R12 =9×109×(4×10−6)2(6×10−2)2 =10N F2=kQ1Q2R22 =9×109×(4×10−6)2(8×10−2)2 =36064=5.62N Resultant force at the right angle vertex is, F=√(F12+F22) =√102+5.622 =√131.56 =11.5N

Three equal charges, each having a magnitude of $4 μ C$ , are placed at the three corners of a right-angled triangle of sides $6 c m, 8 c m$ and $10 c m .$ The force on the charge at the right-angle corner will be