A long cylindrical shell (radius = 2.0 cm) has a charge uniformly distributed on

Question

A long cylindrical shell (radius = 2.0 cm) has a charge uniformly distributed on its surface. If the magnitude of the electric field at a point 8.0 cm radially outward from the axis of the shell is 85 N/C, how much charge is distributed on a 2.0-m length of the charged cylindrical surface? I tried finding the Q value of the first one then doing new/old but that was definitely not right? How do I do this, explain please.

Explanation / Answer

A cylindrical shell of radius 7.00 cm and length 260 cm has its charge uniformly distributed on its curved surface. The magnitude of the electric field at a point 20.0 cm radially outward from its axis (measured from the midpoint of the shell) is 36.0 kN/C. (a) Use approximate relationships to find the net charge on the shell. (b) Use approximate relationships to find the electric field at a point 4.00 cm from the axis, measured radially outward from the midpoint of the shell. Hello pitercelo, in this case we have to use the expression E = (1/2pi epsilon not) Lambda/ r.---------->(1) epsilon not - the permittivity of free space. Where Lambda is the linear charge density. r - the distance from the axis of the charged cylinder. To know about the charge Q on the cylinder we have to use the expression Q = lambda* L (L - length of the cylindrical shell) From (1) lambda = 2* pi *epsilon not* E* r Hence the required charge Q = 2* pi *epsilon not* E* r* L Now conversion is needed for cm to be in metre and kN is to be in N. Plugging the values you will get the answer as 1.0409 micro coulomb b) But in case of finding the field at a point inside the charged cylinder, definitely the field is zero. This is because the point is at a distance of 4 cm from the axis, where as the radius is 7 cm.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.