# What is the electric field strength at the vertex of an equilateral triangle with sides of 0.50 m if the charges at the other vertices are...+2.0 x 10^-5 C.

*print*Print*list*Cite

### 1 Answer

The electric field at point in a distance of r due to a point charge q is given by the following equation,

E = (kq)/r^2

Where k is Coulomb's constant = 9 x 10^9 Nm^2/C^2

The electric field due to one 2 x 10^-5 C charge at the vertex is ,

E = (9x10^9 x 2 x10^(-5))/0.5^2 = N/C

E = 720000 N/C = 720 kN/C

The direction is from that charge to vertex.

The electric field due to the other charge is having the same magnitude 720 kN/C, but the direction is different. The angle between the two fileds are 60 degrees.

We have to find the resultant field using the vector addtion,(using the cosine law or vector parallelogram)

E = sqrt(720^2+720^2 - 2x720x720xcos(180-60))

E = 1247 kN/C

**The resultant electric filed is 1247 kN/C and it's direction is inthe direction of median through the vertex , but in the outward direction.**

**Sources:**