Question

ABC is an equilateral triangle of side 10cm a charge of 6 micro coulomb and 3 microcoulomb are placed at a and b calculate electric field intensity at C

Answer 1

Answer:

ABC is an equilateral triangle of side 10m.

 D is the midpoint of BC.

Charge on B = +100µC ; Charge on C = -100µC ; Charge on D = +75µC

Consider,  put A at the top, D directly below.

B and C have equal charge;

lets find the magnitude of each of those forces.

F = kQq / d² = 9e9 N·m²/C² * 1e-6C * 100e-6C / (10m)² = 9 mN

The vertical components of these two forces (A-B and A-C) cancel; the horizontal forces add:

Fh = 2 * 9mN * cos60º = 9 mN to the right.

For D, the distance A-D is d = 10m * √3/2, so d² = 75 m², and

F = 9e9 N·m²/C² * 1e-6C * 75e-6C / 75m² = 9 mN up

So, the net force on A is

F = √(9² + 9²) mN = 9√2 mN = 9√2e-3 N

Explanation:

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