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
Answer:
The current in the circuit must be zero.
Explanation:
In a RC circuit, the steady state is reached when either the capacitor is fully charged or fully discharged. In either case, there must not be any current through the circuit because if it exists, it will deliver charge to the capacitor and thus change its charge, which is not a steady state.
Answer:
Explanation:
There is no set way to do this. All you have to do is define left and right. Left will be minus and right will be the opposite --- plus.
That is completely arbitrary. It could be the other way around. It does not matter.
Left is minus so: - 600 N is the force going left.
Right plus so: + 500 N
Now just add.
Net Force = +500 - 600
Net Force = - 100 N
So the Net Force is - 100 N going to the left.
Answer:
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