# Current in a cylinder moving with uniform velocity

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. 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.

• 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.
– nasu
Jul 14, 2021 at 15:56
• 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 Jul 15, 2021 at 7:23
• 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.
– nasu
Jul 15, 2021 at 18:07