Short Answer 
The problem involves three +1 ¬¨¬µC point charges at the vertices of a 10 cm equilateral triangle. Each charge generates an electric field of 8.99 ‚àöo 10‚ÄöA¬µ N/C, but due to symmetry, these fields cancel out at the triangle’s center, resulting in a net electric field of 0 N/C.
Step 1: Identify the Geometry and Charges
The first step is to understand the setup of the problem. You have three point charges, each with a charge of +1 µC, positioned at the vertices of an equilateral triangle. The side length of the triangle is 10 cm. This configuration is crucial as it dictates how the electric fields from each charge will interact.
Step 2: Calculate the Electric Field from a Single Charge
Next, calculate the electric field produced by one charge at a vertex. The formula for the electric field (E) from a single charge (q) at a distance (r) is: E = k * q / r¬¨‚â§. Here, k is Coulomb’s constant (8.99 ‚àöo 10‚ÄöAœÄ N m¬¨‚â§/C¬¨‚â§) and the distance (r) to each charge is 0.1 m (converted from 10 cm). By substituting the values, you can compute the electric field:
- Calculate E = 8.99 √o 10‚Aµ N/C for one charge.
Step 3: Determine the Resultant Electric Field
Finally, analyze how the three electric fields interact. Since the system is symmetric, the electric fields from the three charges will point away from the charges and combine. Due to the symmetry of the equilateral triangle, these fields will cancel each other out, resulting in a net electric field of 0 N/C at the center of the triangle. This illustrates the principle of superposition in electric fields.