Explore IDNLearn.com's extensive Q&A database and find the answers you're looking for. Ask your questions and get detailed, reliable answers from our community of knowledgeable experts.
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. :)
We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. Find precise solutions at IDNLearn.com. Thank you for trusting us with your queries, and we hope to see you again.