Question

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?

Short answer

The magnitude of the net electric field these two particles produce at the third corner is.$3.6\times {10}^2\text{N/C}$

Step-by-step solution

The given data

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

Understanding the concept of the electric field

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)

Calculation of the net electric field

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.$3.6x{10}^2\text{N/C}$