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Sagot :
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
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. :)
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