The three point charges 8nC, 6nC & 10 nC are placed at corners of an equilateral triangle 1m. Calculate resultant electric field at centroid of triangle?

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Answer

And -4nC are placed at the vertices of an equilateral triangle with sides of length 2m. Determine the electric field at the fourth vertex.

To find the electric field at the fourth vertex, we need to calculate the electric field contribution from each of the three charges and then sum them up.

Let's denote the charges as q1 = 8nC, q2 = 6nC, and q3 = -4nC.

The electric field at the fourth vertex (P) due to q1 can be calculated using Coulomb's Law:

E1 = k * q1 / r1^2

where k is the electrostatic constant (9 x 10^9 Nm^2/C^2) and r1 is the distance between q1 and P. Since the triangle is equilateral, r1 = 2m.

E1 = (9 x 10^9 Nm^2/C^2) * (8 x 10^-9 C) / (2m)^2
E1 = 18 x 10^3 N/C

Similarly, we can calculate the electric field at P due to q2 and q3:

E2 = (9 x 10^9 Nm^2/C^2) * (6 x 10^-9 C) / (2m)^2
E2 = 9 x 10^3 N/C

E3 = (9 x 10^9 Nm^2/C^2) * (-4 x 10^-9 C) / (2m)^2
E3 = -6 x 10^3 N/C

Now, we can find the total electric field at P by summing up the individual electric fields:

E_total = E1 + E2 + E3
E_total = (18 x 10^3 N/C) + (9 x 10^3 N/C) + (-6 x 10^3 N/C)
E_total = 21 x 10^3 N/C

Therefore, the electric field at the fourth vertex is 21 x 10^3 N/C.

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