A proton and an electron form two corners of an equilateral triangle of side length $\mathbf{2}\mathbf{.0}\mathbf{}\mathbf{\times }\mathbf{}{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathit{m}$.What is the magnitude of the net electric field these two particles produce at the third corner?

The side length of the equilateral triangle,$a=2.0x{10}^{-6}\text{m}$

We denote the electron with subscript, e and the proton with, p. From the figure below we see that

$\mathbf{\left|}{\overrightarrow{\mathbf{E}}}_{\mathbf{e}}\mathbf{\right|}\mathbf{=}\mathbf{}\mathbf{\left|}{\overrightarrow{\mathbf{E}}}_{\mathbf{p}}\mathbf{\right|}$

where, d = 2.0 x 10^–6 m. We note that the components along the y axis cancel during the vector summation. With k = 1/4πε_o and θ = 60^o, the magnitude of the net electric field is obtained as follows:

Formula:

The magnitude of x-component of the electric field on a particle due to an electron or a proton is given as: $\left|E_x\right|=\frac{e\cos \theta }{4\pi {ϵ}_o{d}^2}$ (i)

The magnitude of the net electric field is obtained using equation (i) as follows: (as all vertical components cancel each other from the figure:

$\begin{array}{c}\left|{\overrightarrow{E}}_{net}\right|=2\left|E_x\right|\\ =2\left(\frac{e}{4\pi {ϵ}_o{d}^2}\right)cos\theta \\ =2(8.99\times {10}^9{\text{N.m}}^{\text{2}}{\text{/C}}^{\text{2}})\frac{(1.6\times {10}^{-19}\text{C})}{{(2.0\times {10}^{-6}\text{m})}^2}cos{60}^o\\ =3.6x{10}^2\text{N/C}\end{array}$

Hence, the value of the electric field is.$3.6x{10}^2\text{N/C}$