Question

ABC is an equilateral triangle of side 10 m and D is the
mid-point of BC. Charges of +100,- 100 and +75 uc
are placed at B, C and D respectively. Find the force on
a +1 °C charge placed at A. A
Ans. 9/2 x 10-3 N, at 45° to a line parallel to BC.​

Answer 1

Answer:

At the point A

net field along x – axis

E x = EB cos 60° + EC cos 60°

E y = ED + EB sin 60° – EC sin 60°

Given that a = 10m

QA = + 1 µC

QB = 100 µC

QC = – 100 µC

QD = 75 µC

k = 9*109 N-m2/C2

Electric fields (magnitude) are

So, Ex = 9000 cos 60° + 9000 cos 60°

=

E y = 9000 + 9000 sin 60° – 9000 sin 60°

= 9000 N/C

So net field at point A.

So net force acting on point charger of (+1 µ C) is given by

F = q E n

= (1 × 10– 6) 12727.9

F = 12.73 × 10– 3N

Answer 2

Answer:

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ur ans is,

kindly see the attachment

Explanation:

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