At 4:40, we have Q1-Q2=something, Q1*Q2=something. Now, we know that, // (Q1+Q2) square = (Q1-Q2) square + 4*Q1*Q2 // We now have the value of both, (Q1+Q2) & (Q1-Q2). Q1 & Q2 can be determined easily! Simple calculation!
Sir, great to see your lectures ...you are awesome........... Sir actually this method involves a lot of calculations....instead we can find the ratio b/n charges ie Q1*Q2=4Q3^2.... As we know when the charges are touched .charges ae transferred till they are of same charges....ie Q1=Q3+x and Q2 = Q3-x now we can equate and finally get the answer...... THANK YOU SIR !
Sir, first of all thank you for your videos even in 2021. I have a remark though, wouldn't be simpler if we've just express Q1 as Q2/C1? The we can substitute this into the equation of (Q1-Q2)^2=4C2. We get a relief of a little bit shorter algebra, and since we do not lose any roots the solution will be exactly the same. Can it be a mistake to suppose that either Q1 or Q2 cannot be zero?
Sir, if the negative charge has a greater magnitude than the positive the remainder would contain the negative charges, right?? because then it would be Q2-Q1. Also, when two charges are brought in contact is it induction....when the charges are neutralized?
thank you sir, your videos are more helpful. I was just less in engineering physics, few days after watching your videos, i totally improved. but sir where I can get your PDF notes,I hope it will help me more. thank you.
Some of the excess positive charge of Q1 cancelled out the negative charge of Q2. The remainder positive charge of Q1 is then equally distributed among both 1 and 2
Q1-Q2 is the remaining charge after they are added together. Then the remaining charge is divided equally between the two object (each getting half of the net charge that is left).
Sana, Sometimes the equation becomes "messy" and it is difficult to see what to do next. Replacing a more complicated part of the equation with a simple substitution makes it easier to complete the problem. You just have to re-substitute the original expression back in when you are done
At 4:40, we have
Q1-Q2=something,
Q1*Q2=something.
Now, we know that,
// (Q1+Q2) square = (Q1-Q2) square + 4*Q1*Q2 //
We now have the value of both, (Q1+Q2) & (Q1-Q2).
Q1 & Q2 can be determined easily!
Simple calculation!
Yes i did it in the same way🤗
Sir, great to see your lectures ...you are awesome...........
Sir actually this method involves a lot of calculations....instead we can find the ratio b/n charges ie Q1*Q2=4Q3^2....
As we know when the charges are touched .charges ae transferred till they are of same charges....ie Q1=Q3+x and Q2 = Q3-x
now we can equate and finally get the answer......
THANK YOU SIR !
Sir, first of all thank you for your videos even in 2021. I have a remark though, wouldn't be simpler if we've just express Q1 as Q2/C1? The we can substitute this into the equation of (Q1-Q2)^2=4C2. We get a relief of a little bit shorter algebra, and since we do not lose any roots the solution will be exactly the same. Can it be a mistake to suppose that either Q1 or Q2 cannot be zero?
Sir,if Q2 is also a positive charge but with a lesser magnitude than Q1 then what will Q3 be equal to?Will it be Q1+Q2/2?
This was so helpful! 🙌 thanks you the man
Sir, if the negative charge has a greater magnitude than the positive the remainder would contain the negative charges, right?? because then it would be Q2-Q1.
Also, when two charges are brought in contact is it induction....when the charges are neutralized?
Why when finding out F1, you multipy Q1 and Q2. But for F2, you are squaring Q3? I thought you had to multiply the two q's( Q1 and Q3) to find F2.
I thought so and want about the direction of Q3 and Q1?
You didn't add 5 of 5 into the playlist!
Mushyrulez That is right. I had forgotten about that video. Thanks
Sir i am belongs to india🇮🇳. And I just watch this vedio totally change my mind 🙏🙏
Welcome to the channel!
thank you sir, your videos are more helpful. I was just less in engineering physics, few days after watching your videos, i totally improved. but sir where I can get your PDF notes,I hope it will help me more. thank you.
At this time, there are no PDF notes available.
Oooh!, no matter though it would help me. I will keep following your videos. Fortunately I will know more. May I have your email address?
When you will be adding the fifth video?
could you please explain the neutralizing q2 part why did you minus and not plus and why are you neutralizing q2
please if possible explain in simple terms
Some of the excess positive charge of Q1 cancelled out the negative charge of Q2. The remainder positive charge of Q1 is then equally distributed among both 1 and 2
Hi, thank you so mach for the video!
you do (Q1-Q2)/2=Q3 because one of the charges is negative and the other one is possitive?
❤❤❤❤❤
thanks , but why (Q1_Q2)/2=Q3???
Q1-Q2 is the remaining charge after they are added together. Then the remaining charge is divided equally between the two object (each getting half of the net charge that is left).
When the video is at 6:30, you are adding 4C2( 4 c subscript 2) + 2C1 (2 c subscript 1)= C3 (c subscript 3), how?
Sana,
Sometimes the equation becomes "messy" and it is difficult to see what to do next.
Replacing a more complicated part of the equation with a simple substitution makes it easier to complete the problem.
You just have to re-substitute the original expression back in when you are done