Question

1. Charges of + 5 UC, + 10 UC and - 10 uC are placed in air at
the corners A, B and C of an equilateral triangle ABC, having
each side equal to 5 cm . Determine the resultant force on
the charge at A.​

Answer 1

Answer:

$\\\tt 1.8 \sqrt{3} \times 10^{-6} N$

Explanation:

Now let the following corners have following charges,

$\\\tt A= 5 \times 10^{-9} C$

$\\\tt B= 10^{-8} C$

then

$\\\tt C= -10^{-8} C$

Now,

Force on A because of B =

$\\\tt F_{1} = \dfrac{1}{4 \pi \epsilon_{\circ}} \times \dfrac{5 \times 10^{-9} \times 10^{-8}}{0.05^2} \\=1.8 \times 10^{-6}N$

Now Force on A because of C:

$\\\tt F_{2} = \dfrac{1}{4 \pi \epsilon_{\circ}} \times \dfrac{5 \times 10^{-9} \times (- 10^{-8}) }{0.05^2} \\=-1.8 \times 10^{-6}N$

Since,

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

Resultant force on $\\\tt F_{1}$ and $\\\tt F_{2}$

$\\\tt F_{net} =\sqrt{F_{1}^2 + F_{2}^2 +2 F_{1}F_{2} \cos \theta} \\=\sqrt{2F_{1}^2 - 2F_{1}^2 \cos{120^{\circ}}} \\=\sqrt{2F_{1}^2 - 2F_{1}^2 (\dfrac{-1}{2}) } \\=F_{1} \sqrt{3}$

Putting in values:

$\\\tt 1.8 \sqrt{3} \times 10^{-6} N$