Answer :
ANSWER:
0.09 N
STEP-BY-STEP EXPLANATION:
Given:
Charge A (qA) = 2.5 × 10^-7 C
Charge B (qB) = -2.5 × 10^-7 C
Charge C (qC) = 1 × 10^-7 C
Distance (d) = 5 cm = 0.05 m
The first thing is the electric force between AC and BC, using Coulom's law, just like this:
[tex]\begin{gathered} F_{12}=k\cdot\frac{q_1\cdot q_2}{d^2} \\ \\ \text{ We apply in each case:} \\ \\ F_{AC}=k\cdot\frac{q_A\cdot q_C}{d^2}=9\times10^9\cdot\frac{\left(2.5×10^{-7}\right)\left(1×10^{-7}\right)}{(0.05)^2}=0.09\text{ N} \\ \\ F_{BC}=k\cdot\frac{q_B\cdot q_C}{d^2}=9\times10^9\cdot\frac{(-2.5×10^{-7})(1×10^{-7})}{(0.05)^2}=-0.09\text{ N}=0.09\text{ N} \end{gathered}[/tex]The resultant force due to the two forces is calculated using the following formula applied to the following angle, thus:
[tex]F=\sqrt{(F_{AC})^2+(F_{BC})^2+2\cdot F_{AC}\cdot F_{BC}\cdot\cos\theta}^[/tex]We replace each value in order to calculate the electrical force on the charge at C due to the other two charges:
[tex]\begin{gathered} F=\sqrt{\left(0.09\right)^2+\left(0.09\right)^2+2\cdot0.09\cdot0.09\cdot\:\cos120°} \\ \\ F=\sqrt{0.0081+0.0081-0.0081} \\ \\ F=\sqrt{0.0081}=0.09\text{ N} \end{gathered}[/tex]The electric force on the charge at C due to the other two charges is 0.09 N.
