Question

The drawing shows an equilateral triangle, each side of which has a length of 2.00 cm. Point charges are fixed to each corner, as shown. The 4.00 microC charge experiences a net force due to the charges qA and q B. This net force points vertically downward and has a magnitude of 405 N. Determine the magnitudes and algebraic signs of the charges qA and q B.

Answer 1

Answer:

$-2.60097\times 10^{-6}\ C$

Explanation:

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

r = Distance between charges = 2 cm

The electric force is given by

$F=\dfrac{k4\times 10^{-6}q_A}{r^2}$

Also

$F=\dfrac{k4\times 10^{-6}q_B}{r^2}$

The sum of the force is given by

$\sum F=\dfrac{k4\times 10^{-6}q_A}{r^2}cos 30+\dfrac{k4\times 10^{-6}q_B}{r^2}cos 30\\\Rightarrow \sum F=2\dfrac{k4\times 10^{-6}q_A}{r^2}cos 30\\\Rightarrow q_a=\dfrac{\sum Fr^2}{2\times k4\times 10^{-6}cos30}\\\Rightarrow q_a=\dfrac{405\times 0.02^2}{2\times 8.99\times 10^9\times 4\times 10^{-6}cos30}\\\Rightarrow q_a=2.60097\times 10^{-6}\ C$

Thus $q_a=q_b=-2.60097\times 10^{-6}\ C$.

The negative sign is because the force points downward which is taken as negative