Question

1. [10 points] Use integration to derive the formula for the horizontal component E of the electric field, evaluated at the origin, along the axis of a hollow cylindrical roll of paper (the cylinder has no end caps). The paper has a surface charge density of σ (given) in C/m2, and extends from the origin to a length of L, as shown in the figure below: AY hollow paper cylinder with surface charge density-ơ radius = R E (0,0) -? 0 2L lengths L int: You'll most likely need the formula for the electric field of a one-dimensional ring, as opposed to a zero-dimensional point. If the ring's axis is the x-axis, the ring's electric field formula along the axis is: 1 xQring Ering-4πεο (x2 + a2) 12 where Qring is the charge of the ring, a is the radius of the ring, x is the distance from the center of the ring to the test charge, and the unit vector 1 indicates the direction of this E field. Make sure to explain/show in detail all of the key steps of the setup, since at least 80% of the marks in this type of question are for the setup steps, not the actual integral. You should show your choice for slicing into charge elements, shade in the non-specific charge element you chose to work on, find the formula for the charge element, find the formula for the distance to the charge element, draw the FBD, and use the correct symbols/variables when substituting to set up the integral and limits of integration.

Answer 1

Consider a hollow cylinder whose length is dx and charge dQ which is like a ring let surface charge density = sigma Electric field due to ring of charge Q and distance d is given by Ex^vector = 1/4pi epsilon_0 Qd/(d^2 + r^2)^3/2 Applying this to ring of cylinder we get dE_x = 1/4pi epsilon_0 dQ/(x^2 + R^2)^3/2 times x equation (1) \r\n As sigma = dQ/dA = Q/A therefore dQ = sigma dA or dQ = sigma(2 pi R) dx putting this value of d in eq(1) d Ex = 1/4 pi epsilon_0 sigma(2 pi R) x dx/(x^2 + R^2)^3/2 equation (2) Integrating eq, (2) between x = 0 to L E_x = integral^L_0 dEx = sigma(2 pi R)^L/4pi epsilon_0 integral^L_0 x/(x^2 + R^2)^3/2 dx E_x = sigma(2pi R)/4pi epsilon_0 Vertical-bar (-1/(x^2 + R^2)^1/2) Vertical-bar ^L_0 E_x = -sigma(2pi R)/4pi epsilon_0 times (1/squareroot L^2 + R^2 - 1/(R^2)^1/2) \r\n As sigma = Q/2pi RL therefore sigma (2pi R) = Q/L therefore