Question

If the charge q is placed at the centre of the line joining two equal charges Q such that the system is in equilibrium then the value of q is? 
Really need it, with a good explanation, will appreciate such answer. ​

Answer 1

Answer:

q is 0

Explanation:

Now another charge q is placed at the middle point of the line and the system is in equilibrium, that means net force on charge q is 0

Answer 2

Let's just suppose this situation, so it's like this;

Q --------------- q --------------- Q

Now, acc. to the question, the system is equilibrium which means that possibly both the Q should be exerting some pressure on the middle charge q, so let's suppose Q be +ve and q be -ve

The charge that we have kept at a certain distance something like this;

+Q --------------- -q --------------- +Q

<––– d ––– > <––– d ––– >

Now, we +Q will be exerting some force on the another +Q that it should be acc. to columbs law like this;

F = k × Q1Q2/r²

F = k × QQ/(2d)²

F1 = k × Q²/4d²

Also, when this +Q will exert force on -q, it could be written as;

F = k × Q1Q2/r²

F2 = k × Qq/d²

From the question, we know that the system is a equilibrium which means that these two force F1 and F2 will be equal to each other;

= k × Q²/4d² = k × Qq/d²

• Eliminate the common values

= Q/4 = q

• Wait wait! The answer had still not came! remember this q is actually -q?

= Q/4 = -q

= -Q/4 = q

So, the charge on q is -Q/4. :)