Question

three charges 5uc and 10uc and –5uc are fixed at corner of an equilateral triangle of side 5 cm . find the coulomb force on A due to B and C​

Answer 1

Given ,

Three charges 5 uc and 10 uc and -5 uc are fixed at corner of an equilateral triangle of side 5 cm

We know that , the force b/w two charges is given by

$\boxed{ \tt{F = k \frac{ q_{1}q_{2} }{ {(r)}^{2} } }}$

Thus ,

Force on A due to B will be :

$\tt \implies F_{AB} = \frac{9 \times {(10)}^{9} \times 5 \times {(10)}^{ - 6} \times10 \times {(10)}^{ - 6} }{ {\{ 5 \times {(10)}^{ - 2} \}}^{2} }$

$\tt \implies F_{AB} = \frac{45 \times 10 \times {(10)}^{ - 3} }{25 \times {(10)}^{ - 4} }$

$\tt \implies F_{AB} = \frac{45 \times 2 \times 10}{5}$

$\tt \implies F_{AB} =180 \: \: newton$

Force on A due to C will be :

$\tt \implies F_{AC} = \frac{9 \times {(10)}^{9} \times 5 \times {(10)}^{ - 6} \times 5 \times {(10)}^{ - 6} }{ { \{5 \times {(10)}^{ - 2} \} }^{2} }$

$\tt \implies F_{AC} = \frac{9 \times 25 \times {(10)}^{ - 3} }{25 \times {(10)}^{ - 4} }$

$\tt \implies F_{AC} =90 \: \: newton$

Now , the magnitude of resultant force will be

$\tt \implies F = \sqrt{ {(90)}^{2} + {(180)}^{2} + 2 \times 90 \times 180 \times \cos(120) }$

$\tt \implies F = \sqrt{8100 + 32400 + 2 \times 90 \times 180 \times ( - \frac{1}{2} )}$

$\tt \implies F = \sqrt{8100 + 32400 - 16200}$

$\tt \implies F = \sqrt{24300}$

$\tt \implies F = 115.8 \: \: newton$

The force on A due to B and C is 115.8 N