91影视

As given in Fig., particle 1 $(q_1=+2.00\text{pC})$, particle 2$(q_2=鈭�2.00\text{pC})$, and particle 3 $(q_3=+5.00\text{pC})$form an equilateral triangle of edge length.$a=9.50\text{鈥塩尘}$

Using the concept of Coulomb's law of electrostatics, the value of the net force on the third particle can be calculated. Similarly, using a similar concept, we can see that the electric field lines from two equal and oppositely charged bodies cancel each other, resulting in a net-zero force component of the third particle.

Formula:

The magnitude of the electrostatic force acting on a particle 1 due to particle 2 that is making an angle with each other, $F_1=\frac{q_1{q}_{2}cos胃}{4蟺{蔚}_o{a}^2}$ (i)

From symmetry, we see the net force component along the y-axis for two charges of equal magnitude and opposite direction is zero.

The net force component along the x axis points rightward. With 胃 = 60^o, the magnitude of the net force acting on particle 3 due to particle 1 and particle 2 using equation (i), is given as:

$\begin{array}{c}{F}_3=2\frac{q_3{q}_{1}cos胃}{4蟺{蔚}_o{a}^2}\text{(}鈭�\left|q_1\right|=\left|q_2\right|\text{)}\\ F_3=\frac{k{q}_3{q}_1}{a^2}\text{(}鈭�{\text{cos(60}}^0\text{)=1/2)}\\ =\frac{(8.99脳{10}^9\text{N}鈰�\frac{\text{m}}{{\text{C}}^{\text{2}}})(5.00脳{10}^{鈭�12}\text{C})(2.00脳{10}^{鈭�12}\text{C})}{{\left(0.0950\text{m}\right)}^2}\\ =9.96脳{10}^{鈭�12}\text{N}\end{array}$

Hence, the value of the net force is.$9.96脳{10}^{鈭�12}\text{N}$