Question

5) Find the Electric field strength at point A. Indicate direction with an arrow

Answer 1

`||\vec E_A||=1.11446 * 10^6 \ (N)/(C)`, where the vector arrow will point from the charge towards point A.

Conceptual:

What is an electric field?

An electric field is a physical field produced by charged particles, these electric fields have the ability to exert forces on other charged particles. The following formula can be used to find the electric field (as a vector) at a point in space. "k_e" is Coulomb's constant and "
`\hat r`" indicates the direction vector that point from the charge towards the field you are trying to calculate.

`\boxed{\left\begin{array}{ccc} \text{\underline{Equation for Electric Field:}} \\\\\ \vec E=(k_eq)/(r^2) \hat r \\k_e=8.99 * 10^9(Nm^2)/(C^2) \end{array}\right }`

Step-by-step explanation:

Given:

`q=6 * 10 ^(-6) \ C\\\\r=0.22 \ m`

Find:

`\vec E_A=?? \ (N)/(C)`

`\vec E_A=(k_eq)/(r^2) \hat r\\\\\Longrightarrow \vec E_A=((8.99 * 10 ^9)(6 * 10 ^(-6)))/((0.22)^2) \cdot( < 0,-0.22 > )/(√((0)^2+(-0.22)^2) ) \\\\\Longrightarrow \vec E_A= 1.11446 * 10^6 \cdot < 0,-1 > \\\\\Longrightarrow \vec E_A= < 0,-1.11446 * 10^6 > (N)/(C) \\\\\Longrightarrow||\vec E_A||=√((0)^2+(-1.11446 * 10^6\))^2) \\\\\therefore \boxed{\boxed\vec E_A}`

Thus, the electric field strength at point A is found. The vector arrow will point from the charge, q, towards point A.