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This question might seem silly. I solved this question in the following way. Although, The answer I got from this method is correct but I still have a conceptual doubt in my solution. enter image description here I have put $\frac{dx}{dt}=v$ but here $dx$ is just signifying the length of the small element I took. So how can I put this equal to the velocity of the cylinder. According to me when the cylinder moves forward, $dx$ element must also move forward so $dx$ must remain constant.
Where am I wrong? If this solution is wrong, please tell the correct way to solve this question.

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  • $\begingroup$ You have surface charge density and not linear chare density. It is a cylinder, not a thin wire. The formula for electric field is not appropriate for your problem. $\endgroup$
    – nasu
    Jul 14, 2021 at 15:56
  • $\begingroup$ no. the formula is correct. if you convert the surface charge density into linear charge density, you will get the same relation as we got for a line charge $\endgroup$
    – Nimit Jain
    Jul 15, 2021 at 7:23
  • $\begingroup$ Yes, you are right. And dx here is the witdth of the cylindrical ring that passes any point along the axis during the time dt. So you have dx=vdt. To calculate the charge on this ring you use the area of the ring which depends on dx. $\endgroup$
    – nasu
    Jul 15, 2021 at 18:07

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  1. I Think this is a good question, because you have arrived at the right answer but do not feel confidence in your methodology. So well dome so far!
  2. I think you are right to feel uneasy about the method, because there are two different x-coordinates involved, in principle. The first dx is an increment in a coordinate system fixed in the cylinder (separation of two points A and B along the cylinder axis); the second dx is an increment in a coordinate fixed in the laboratory (motion of one fixed point C in a time delta t). So at the very least you are recycling a symbol without re-defining it. It turns out to be benign because you are in effect choosing the point C to be A at time t=0, and the time delta t to be the time it takes A to arrive the point occupied by B at t=0.
  3. As to the point raised by nasu, that depends how you are conceptualising the symbol lambda. I read it as the charge per unit length, which is the circumference times the charge per unit area, and in the equation for E you are dividing put the circumference again, to get the surface charge and hence E.But nasu read it as a line cherge, that is zero cross-section, which is indeed geometrically wrong.

So one message is: the words to describe the symbols are really important!

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