Find the net force acting on charge 'Q'
Answers
The sum is all about finding symmetry !
I would recommend you to solve these problems considering and finding symmetry as much as possible.
For example in this sum
Because the two charges positive q here are producing force on Q equal in magntiude and opposite in direction placed at r/2 the total force produced due to them on the charge Q will be zero.
Now, the only forces that'll produce an impact on the charge Q are the ones by charges q placed distantly.
Well that distance can be calculated as √r^2 + r^2/4 = √5r/2 units
So, force due to one of those will be
F = k Qq/(√5r/2)^2 = 4kQq/5r^2
Similar force would be applied by another q that's placed right below the previous q charge.
So, same force of 4kQq/5r^2 will be produced.
Now, by some symmetry and geometric analysis you can see that the forces are acting perpendicularly on charge Q
Now, their net resultant force would be √2 F = √2 kQq (4)/5r^2
Where k = 4π€(0)
Hope this helps you !
If something is unclear, do tell me down in the comment section.