Problem 19.13

19.13. a)
By symmetry, the point where the electric field is zero is at the center of the triangle.

b)
At point P, the magnitude of the contribution from each of the charges to the electric field is given by
$E = \frac{kq}{a^{2}}$
We want to split this into components. By symmetry, we can see that the $\hat{x}$ component is zero. So, to find the $\hat{y}$ component, we will take twice the $\sin 60{}^{\circ}$.
$E_{y} = \frac{kq}{a^{2}} (2 \sin 60{}^{\circ})$
$E_{y} = 1.73 \frac{kq}{a^{2}}$
So, the total electric field is given by: $\vec{E}= 1.73 \frac{kq}{a^{2}} \hat{j}$