Doubt in Physics

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Koi solve karo na mera pho rsquare/12episilon not aa raha hai
Ans D hai

@Cooljet123 @Naman1 @mihir @Aswin24 @Ishaan_Shrivastava @harsh2 bro aur jo bhi online hai
@Aastik_Guru jab aana to bana dena ye

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Mereko plus sign/minus sign Ka Nahi pata but if this helps then like kardena @Vibhu_Kumar

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@Ishaan_Shrivastava bhai mai kaha galti kar rahu hu bata na :no_mouth::frowning:

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Bro I am not able to follow your solution at all… Can you write the expression for electric field that you used, again? @Vibhu_Kumar

Like I can see a bunch of stuff that you might have used but there is a lot of rough calculation which is a bit confusing

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I am getting this answer…

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Dekho mai aise kiya ki A pe nikala bada wala aur sphere aur chota wala sphere dono ka madad se (bada-chota)

Same I used for B

Mai E=kq/r2 outside aur pr/3Eo andar jo formula hota hai uska istamal kiya

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Electric field inside the sphere is KQr/R^3

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Bro p me convert karega to wahi aayega Pr/3Eo

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Okay. And using this, what values of absolute potential did you get for both A and B?

Oh Haan cooljet waala formula hi use karlo. But I am not able to understand the mistake you made, as long as you integrated properly

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Yeah, got what you said. Use potential at a point inside solid charged sphere : \frac{KQ}{2R^3} (3R^2 -r^2)

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Maine net change in electric field naikal aur fur v=e.dl ki madad se change in potential nikala jisme l =r/2 liya dono ke beech ka ( A -B)length

Ab samjaha do mara galat ku ho raha hai
@Ishaan_Shrivastava @Cooljet123

Integrate electric field directly from points R/2 to R. I don’t think you should subtract the fields…Because field is not constant.

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Ok bro

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Ah so you did ∆E.∆L.

This is wrong for the following reason:

Potential difference is the integral of -E.dl which simply doesn’t equate to what I wrote in first line of my comment.

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Sukriya bhailog @Cooljet123 @Ishaan_Shrivastava

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Bump for the original question

Is the answer 5m?