Three identical charges, each having a value 1.0 × 10–8 C, are placed at the corners of an equilateral triangle of side 20 cm. Find the electric field and potential at the center of the triangle.

Given:
Value of three identical charges: q = 1.0× 10–8 C
Side of the equilateral triangle: l = 20cm = 0.2m

From the diagram,

A,B and C are the three vertices having equal charge q.
EA,EB and Ec are the electric fields at the center of the triangle due to charges A,B and C respectively.
h is the height of the equilateral triangle
and r is the distance from the center of the triangle to it’s all three vertices.

Formula used:
Formula for potential at a point is:

Where k is a constant and k= =9× 109 Nm2C-2 .q is the point charge and r is the distance between the centre of the triangle and the vertex.

Since charges are equal at A,B and C:
The Field from B and C are resolved into horizontal and vertical components as seen from the figure.
Here θ is 30° as every angle of an equilateral triangle is 60o .
The horizontal components balance each other.
Therefore net electric field ,
Enet = EA -(EB sinθ+ECsinθ)
Enet= E-(Esinθ+Esinθ)
Enet = E(1-sin(30)-sin(30) )
Enet =E(1-0.5-0.5)
Enet =0
Thus, the electric field at the center of the given equilateral triangle is zero.
Now, using Pythagoras theorem to find value of h,

We know that, in an equilateral triangle
Thus we get,

Since Electric field is same for all three points: VA=VB=VC
The potential at the center is :

Hence , potential at the centre of the triangle is 2341 V and Electric field at the center is zero.

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