enter image description here

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.

  • $\begingroup$ 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

1 Answer 1

  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!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.