Please answer the following questions
Answers
Okay so thank you for asking question from my favourite section xD Well,the questions I see here are kinda statements which you might find in most of the notes of reputed institutes (except the first one). So,it will be better for you if you memorize these answers so as to save your time during the exam.
The first question is pretty simple, like it doesn't even deserve a place in most of the exams xD but is just asked sometimes just to let a student know that charges are added algebraically. I don't even think I need to calculate it, I can directly see the answer of this question is option (b) but still here we go.
Solution 1 :
We know that the force of interaction between the two point charges is given by :
F = KQ1Q2/r²
- K = 1/4πϵ
- Q1,Q2 = Point charges
- r = Distance between the point charges
Q1 = 6 and Q2 = 2 and it is mentioned that charge of -2 is added to both these charges,so the net charge on Q1 and Q2 respectively will be :
- Q1 + (-2) = 6 - 2 = 4 = q1
- Q2 + (-2) = 2 - 2 = 0 = q2
So,now we need to find the interaction between q1 and q2. Just substitute the values in the formula,
=> F = K × 4 × 0 /r²
=> F = 0
So if -2 charge is added to the charges then the force of interaction between them will become zero.
Solution 2 :
This kind of question really can be exciting to solve at times because it's fun to guess the sign of the charge which one can do mostly using their sense,no calculations required. And if once the sign of charge is known the other options containing sign which isn't going to help in keeping the system in equilibrium can be eliminated,so you will have to work with merely one or two options making your task simpler.
So, since in the question we are having two positive identical charges and we know that like charges repel each other,so we need to bring in a negative charge between the two which will keep the two charges are rest and hence maintaining an equilibrium.
So, according to the situation mentioned in the question,we have a total of three point charges. But Coloumb's law of electrostatic force is valid only for two charges. So,what we do in such questions is like we make pairs and then apply the formula.
So,the electrostatic force between the right extreme charge and the middle charge :
=> F = K Q q/(r/2)²
=> F = KQq/r²/4
=> F = 4KQq/r²
This will be same for the left extreme charge as well.
Now, let's calculate the charge on either of the positive charges due to the negative charge q kept at centre and due to the other positive charge.
Fq = 4KQq/r²
FQ = KQQ/r² => KQ²/r²
Since equilibrium is required,the net force acting on Q must be zero i.e F = 0
=> 0 = 4KQq/r² - KQ²/r²
(Why are we subtracting 4F from F? Because of the distance,since -q charge is kept closer to the charge Q,the impact of -q will be greater in magnitude)
=> 0 = 4KQq - KQ² / r²
=> 0 = 4KQq - KQ²
=> 4KQq = KQ²
=> 4q = Q
=> q = Q/4
And since we know that the sign of the charge is negative according to the situation of equilibrium,so, q = - Q/4.