Each corner of a right-angled triangle is occupied by identical point charges "A", "B", and "C" respectively. Draw a sketch of this arrangement. "A" exerts force F on "B". An equal force F is exerted by "C" on "B" (/_ ABC= 90 degrees). Determine an expression for the net force on "B".

Fnet = F√2

Fnet = kq²/r² √2

Explanation:

A exerts a force F on B, and C exerts an equal force F on B perpendicular to that.  The net force can be found with Pythagorean theorem:

Fnet = √(F² + F²)

Fnet = F√2

The force between two charges particles is:

F = k q₁ q₂ / r²

where

k is Coulomb's constant, q₁ and q₂ are the charges, and r is the distance between the charges.

If we say the charge of each particle is q, then:

F = kq²/r²

Substituting:

Fnet = kq²/r² √2

the answer is b ) square root of 5

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