3 identical charges each having charge 10 coulomb are placed at the corners of an equilateral triangle of side 20cm find the electric field at the centre of triangle
To find the electric field at the center of an equilateral triangle, we can use the principle of superposition.
First, let's calculate the electric field at the center of the triangle due to one charge. The electric field at a point due to a single charge is given by:
E = k * q / r^2
Where:
- E is the electric field
- k is the Coulomb constant (k ≈ 9 × 10^9 Nm^2/C^2)
- q is the charge
- r is the distance from the charge to the point
In this case, the distance (r) between the charge and the center of the triangle is the length of one side of the equilateral triangle, which is 20 cm (or 0.2 m).
Using this value in the equation, we have:
E1 = k * q / r^2 = (9 × 10^9 Nm^2/C^2) * (10 C) / (0.2 m)^2
Now, since the charges are symmetrically arranged at the corners of the triangle, the electric fields due to each charge will cancel each other out along the horizontal and vertical directions. Thus, the only component of the electric field that remains is the vertical component.
To calculate the net electric field at the center of the triangle, we need to consider the electric field vectors due to each charge and add them together. Since the charges are identical, all three electric field vectors will have the same magnitude.
The magnitude of the sum of these vectors is:
E_total = 3 * E1
Substituting the value of E1:
E_total = 3 * [(9 × 10^9 Nm^2/C^2) * (10 C) / (0.2 m)^2]
Calculating this value will give us the electric field at the center of the triangle.