three equal charges are fixed at the howdy corners of a rectangle what is the force on the top left charge from the bottom left

Let the resultant be

A= ( C2 + B2 )1/2

So

sin = B / ( C2 + B2 )1/2

And

cos = C / ( C2 + B2 )1/2

So the total force is

F = FA + FC + FY

= - k y2 i / c2 + k y2 j / B2 + k y Q ( -cos i + sin j ) / ( C2 + B2 )

F = 0

- k y2 i / C2 + k y2 j / B2 + k y Q ( -sin i + sin j ) / ( L2 + H2 ) = 0

- k y2 i / C2 + k y2 j / B2 + k y Q ( - C / ( C2 + B2 )1/2 i + B/ ( C2+ B )1/2 j ) / ( C2 + B2 )1/2 = 0

Q = - y ( C2 + B )3/2 / C3

k y2 / B2 + k y Q B/ ( C2 + B2 )3/2 = 0

Q = -y ( C2 + b2 )3/2 / B

So equating both " Q " values

- y ( C2 + B2 )3/2 / L3 = - y( C2 + B2 )3/2 / B

so the possible length and the relationship is C= B

the answer is b, have negligible mass