Your question has been heard loud and clear.
Well it depends on the magnitude of charges. Generally , when both positive charges have the same magnitude , their equilibrium point is towards the centre joining the two charges. But if magnitude of one positive charge is higher than the other , then the equilibrium point will be towards the charge having lesser magnitude.
Now , a negative charge is placed in between the two positive charges. So , if both positive charges have same magnitude , they both pull the negative charge towards each other with an equal force. Thus the equilibrium point will be where the negative charge is placed because , both forces are equal , and opposite , so they cancel out each other at the point where the negative charge is placed. However if they are of different magnitudes , then the equilibrium point will be shifted towards the positive charge having less magnitude.
Thank you
Answer:
Explanation:
We know that acceleration is change in velocity over time.
v is the final velocity and u is the initial velocity.
Solve for v.
Multiply both sides by t.
Add u to both sides.
To solve this problem we will apply the concepts related to Ohm's law and Electric Power. By Ohm's law we know that resistance is equivalent to,
Here,
V = Voltage
I = Current
While the power is equivalent to the product between the current and the voltage, thus solving for the current we have,
Applying Ohm's law
Therefore the equivalent resistance of the light string is 
To solve this problem, we will get f and then we will use it to calculate the power.
So, for this farsighted person,
do = 25 cm and di = -80
Therefore:
1/f = (1/25) + (1/-80) = 0.00275 = 0.275 m
Power = 1/f = 1/0.275 = +3.6363 Diopeters.
This means that the lens is converging.
