Two particles, each with a charge of magnitude$\mathbf{12}\mathbf{}\mathbf{\text{nC}}$, are at two of the vertices of an equilateral triangle with edge length$\mathbf{2}\mathbf{.0}\mathbf{}\mathbf{\text{m}}$. What is the magnitude of the electric field at the third vertex if (a) both charges are positive and (b) one charge is positive and the other is negative?

Two charged particles of magnitude$Q=12\text{nC}$, are at two of the vertices of an equilateral triangle with edge length,$d=2.0\text{m}$ .

Using the same concept of the electric field, and the condition of positive and negative charges, we can get the net electric field on the third vertex from the other two charges.

Formulae:

The electric field at a point due to a charge,

$\overrightarrow{E}=\frac{q}{4\pi {\epsilon }_o{r}^2}\widehat{r}\text{where,}\widehat{r}=\cos \theta \widehat{i}+\sin \theta \widehat{j}$ (i)

Where, r = The distance of field point from the charge

q = charge of the particle

From symmetry, we see the net field component along the x axis is zero; the net field component along the y-axis points upward. With, $\theta ={60}^o$the electric field due to the two positive charges is given using equation (i) as:

$\begin{array}{c}{E}_{net,y}=2\frac{Qsin\theta }{4\pi {ϵ}_o{a}^2}\\ =\frac{2\times (9\times {10}^9{\text{NC}}^{\text{2}}{\text{/m}}^{\text{2}})(12\times {10}^{-9}\text{C})\times (\sqrt{3}/2)}{{\left(2\text{m}\right)}^2}\\ =46.71\text{N/C}\\ \approx 47\text{N/C}\end{array}$

Hence, the value of the electric field is.$47\text{N/C}$

From symmetry, we see in this case that the net field component along the y axis is zero; the net field component along the x axis points rightward. With, the electric field due to the charges (negative and positive) is given using equation (i) as:

Hence, the value of the electric field is.$27\text{N/C}$