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romanna [79]
3 years ago
8

Consider a long, cylindrical charge distribution of radius R with a uniform charge density rho. Find the electric field at dista

nce r from the axis, where r < R. (Use any variable or symbol stated above along with the following as necessary: ε0.)
Physics
1 answer:
Flauer [41]3 years ago
5 0

Answer:

E = p*r / 2*e_o

Explanation:

Given:

- Volume of cylinder V = pi*r^2*L

- Surface area A = 2*pi*r*L

- permittivity of space : e_o

Find:

Electric field E at distance r from the axis, where r < R.

Solution:

Step 1: Application of Gauss Law

- Form a Gaussian surface within the cylinder with r < R. Th cylinder has two surfaces i.e curved surfaces and end caps. Due to long charge distribution the flux through is zero, since the surface dA of end cap and E are at 90 degree angle to one another; hence, E . dA = E*dA*cos(90) = 0. For the curved surface we have:

              (surface integral) E.dA = Q_enclosed / e_o

Step 2: The charge enclosed (Q_enclosed) is function of r and proportional density:

              Q_enclosed = p*V

              Q_enclosed = p*pi*r^2*L

Step 3: The area of the curved surface:

              dA = 2*pi*r*L

Step 4: Compute E:

              E*(2*pi*r*L) = p*pi*r^2*L / e_o

              E = p*r / 2*e_o

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\boxed{\boxed{x = 0.8\:m}}\end{array}}\qquad\quad\checkmark

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