Answer:![F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]](https://tex.z-dn.net/?f=F_%7Bnet%7D%3D%5Cfrac%7Bkq%5E2%7D%7B%28L%29%5E2%7D%5Cleft%20%5B%20%5Cfrac%7B1%7D%7B2%7D%2B%5Csqrt%7B2%7D%5Cright%20%5D)
Explanation:
Given
Three charges of magnitude q is placed at three corners and fourth charge is placed at last corner with -q charge
Force due to the charge placed at diagonally opposite end on -q charge
where 
Distance between the two charges
negative sign indicates that it is an attraction force
Now remaining two charges will apply the same amount of force as they are equally spaced from -q charge
The magnitude of force by both the charge is same but at an angle of 
thus combination of two forces at 2 and 3 will be
Now it will add with force due to 1 charge
Thus net force will be
-- The vertical component of the ball's velocity is 14 sin(<span>51°) = 10.88 m/s
-- The acceleration of gravity is 9.8 m/s².
-- The ball rises for 10.88/9.8 seconds, then stops rising, and drops for the
same amount of time before it hits the ground.
-- Altogether, the ball is in the air for (2 x 10.88)/(9.8) = 2.22 seconds
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-- The horizontal component of the ball's velocity is 14 cos(</span><span>51°) = 8.81 m/s
-- At this speed, it covers a horizontal distance of (8.81) x (2.22) = <em><u>19.56 meters</u></em>
before it hits the ground.
As usual when we're discussing this stuff, we completely ignore air resistance.
</span>
I'm not exactly sure but I'm thinking that it's the last one. Sorry if I'm wrong
Answer:
The SI units for energy is Joules.
