Question

In Fig. 22-66, particle 1 (of charge$\mathbf{+}\mathbf{1}\mathbf{.00}\mathbf{}\mathbf{\text{μC}}$), particle 2 (of charge), and particle 3 (of charge Q) form an equilateral triangle of edge length a. For what value of Q(both sign and magnitude) does the net electric field produced by the particles at the center of the triangle vanish?

Short answer

The value of for which the net electric field at the center of the triangle vanishes is $+1.00\text{μC}$.

Step-by-step solution

The given data

In Fig., particle 1 (of charge$q_1=1.00\text{μC}$), particle 2 (of charge$q_2=1.00\text{μC}$), and particle 3 (of charge$Q$) form an equilateral triangle of edge length, a.

Understanding the concept of electrostatics

Using the concept of Coulomb's law of electrostatics and the condition of achieving the symmetry for the net electric field to be zero, we can calculate the value of the required charge by canceling their x and y-components.

Formula:

The electric field due to a particle on a point, $E=\frac{q}{4\pi {\epsilon }_0{r}^2}$ (i)

Calculation of the value of the charge, Q

As all the charges are at a distance,$r=a/\surd 3$from the center of the triangle, the net electric field at the center of the triangle is achieved (for) using equation (i) as follows: (If the yaxis is vertical, then (so the electric field due to the two charges is balanced by the electric field of charge, Q)

$\begin{array}{c}2kqsin{30}^o/r^2=kQ/r^2\\ Q=2q\sin {30}^0\\ =q\\ =+1.00\text{μC}\end{array}$

Hence, the value of the charge is.$+1.00\text{μC}$