Question

Three charges of 10mc,5mc, and -5mc are placed at the corners of an equilateral triangle of side 0.1m .
find the resultart force experienced
by 10MC.​

Answer 1

Answer:

$\\\tt \sqrt{3} (45 \times 10^{2}) N$

Explanation:

Consider the given attachment  for diagram.

+vs charges have flux vectors moving away from them and -ve have towards them (convention) .

Here A has $\\\tt 10^{-2}C$

Then B = $\\\tt 5 \times 10^{-2}C$

C= $\\\tt -5 \times 10^{-2}C$

Since angle between AC and AB as shown in the diagram is $\\\tt 120^{\circ}$

Force between A and C be given by:

$\\\tt F_{1} = \dfrac{1}{4\pi\epsilon_{\circ} } \dfrac{10^{-2} \times 5 \times 10^{-3}}{0.1^{2}} \\= 9\times 10^{9} \times 5 \times 10^{-7} N\\= 45 \times 10^{2} N$

Force between A and C be given by:

$\\\tt F_{2} = \dfrac{1}{4\pi\epsilon_{\circ} } \dfrac{10^{-2} \times (-5) \times 10^{-3}}{0.1^{2}} \\= 9\times 10^{9} \times (-5) \times 10^{-7} N\\= -45 \times 10^{2} N$

Since $\\\tt F_{1} = - F_{2}$

Net force by these 2 forces:

$\\\tt \sqrt{F_{1} ^{2} + F_{2}^{2} + 2F_{1} F_{2} \cos\theta}\\=\sqrt{F_{1} ^{2} + F_{1}^{2} - 2F_{1} ^{2} \cos120^{\circ}}\\=\sqrt{F_{1} ^{2} + F_{1}^{2} - 2F_{1} ^{2} (\dfrac{-1}{2}) }\\=\sqrt{3F_{1} ^{2} }\\\\=F_{1} \sqrt{3}$

So the net force experienced by $\\\tt 10mC$

$\\\tt \sqrt{3} (45 \times 10^{2}) N$