Find the electric field a distance r from a line of positive charge of infinite

Question

Find the electric field a distance r from a line of positive charge of infinite length and constant charge per unit length lambda (see figure (a)). Conceptualize The line of charge is infinitely long. Therefore, the field is the same at all points equidistant from the line, regardless of the vertical position of the point in figure (a). We expect the field to become weaker as we move farther away from the line of charge. (a) An infinite line of charge surrounded by a cylindrical gaussian surface concentric with the line of charge. (b) An end view shows that the electric field on the cylindrical surface is constant in magnitude and perpendicular to the surface. Categorize Because the charge is distributed uniformly along the line, the charge distribution has cylindrical symmetry and we can apply Gauss's law to find the electric field. Analyze The symmetry of the charge distribution requires that be perpendicular to the line charge and directed outward as shown in figures (a) and (b). To reflect the symmetry of the charge distribution, let's choose a cylindrical gaussian surface of radius r and length l that is coaxial with the line charge. For the curved part of this surface, E is constant in magnitude and perpendicular to the surface at each point, satisfying conditions (1) and (2) of the following. Condition (1): The value of the electric field can be argued by symmetry to be constant over the portion of the surface. Condition (2): The dot product E . dA can be expressed as a simple algebraic product EdA because E and dA are parallel. Condition (3): The dot product E. dA is zero because E and dA are perpendicular. Furthermore, the flux through the ends of the gaussian cylinder is zero because E is parallel to these surfaces. That is the first application we have seen of condition (3). We must take the surface integral in Gauss?s law over the entire gaussian surface. Because E. dA is zero for the flat ends of the cylinder, however, we restrict our attention to only the curved surface of the cylinder. Apply Gauss's law and conditions (1) and (2) for the curved surface, noting that the total charge inside our gaussian surface is lambda l. Substitute the area A = 2 pi r l of the curved surface: Solve for the magnitude of the electric field. (Use the following as necessary: ke, lambda, and r.) Finalize This result shows that the electric field due to a cylindrically symmetric charge distribution varies as 1/r, whereas the field external to a spherically symmetric charge distribution varies as 1/r^2. The equation for E can also be derived by direct integration over the charge distribution. Consider a long cylindrical charge distribution of radius R = 19 cm with a uniform charge density of p = 12 C/m^3. Find the electric field at a distance r = 33 cm from the axis. Use the charge density to calculate the total charge enclosed by the Gaussian surface, and use Gauss's law to find the field. N/C

Explanation / Answer

consider a gaussian cylinder of radius r and length L

amount of charge inside the cylinder of radius r and length L

qin = p*pi*R^2*L

p = charge density

flux through the cylindrical surface = E*A

A = area of the surface of the cylinder orf radius r and L

flux = E*2*pi*r*L

from gauss law

tota flux q(inside)/eo

E *2*pi*r*L = p*pi*R^2*L/eo

E = p*R^2/2*eo*r

E = (12*0.19^2)/(2*8.85*10^-12*0.33)

E = 7.42*10^10 N/C <<-----answer

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