Question

two charges -1 mc and 2mc are placed at the corners A and B of an equilateral triangle ABC of side 2m.calculate the electric field at c​

Answer 1

Answer:

From charge -1E-6 at point A, field vector v1

magnitude a = 1E-6(9E9/2^2) = 2250 N/C

and direction is C to A (negative charge).

From charge +2E-6 at point B, field vector v2

magnitude b = 2E-6(9E9/2^2) = 4500 N/C

and direction is B to C (positive charge).

For convenience, note that b = 2a.

The angle between sides of the equilateral triangle at C is 60°, but because the vectors CA and BC have opposite directions, their proper separation angle is 180-60 = 120° counter clockwise from BC to CA. It may help to sketch the triangle and vectors.

By law of superposition, the vector sum v1+v2 is the total electric field vector. Addition is easier in Cartesian coordinates. Using line BC as a Cartesian x-axis, the (x, y) vector coordinates of v2 become

v2 = ( b, 0 ) = ( 2a, 0 )

and v1 on line CA is 120 degrees counter clockwise from our arbitrary x-axis, so its (x, y) coordinates become

v1 = ( a*cos120°, a*sin120° )

= ( -0.5a, 0.866a )

The Cartesian total field vector at point C

T = v1+v2 = ( 2a-0.5a, 0.866a+0 )

T = a( 1.5, 0.866 )

Back to polar coordinates, total field vector

T magnitude = a*sqrt(1.5^2 + 0.866^2) N/C

and direction angle θ counter clockwise from x-axis v2 (line BC), where tanθ = 0.866/1.5, so

θ = 30 degrees.

You can also refers to the attachment another answer

Answer 2

Explanation:

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