Explore BrainMass

Explore BrainMass

    Electric field due to a charged disc

    Not what you're looking for? Search our solutions OR ask your own Custom question.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    A uniformly charged disk of radius 35.0 cm carries charge with a density of 7.90 x 10^-3 C/m^2. Calculate the electric field on the axis of the disk at
    (a) 5.0 cm,
    (b) 10.0 cm,
    (c) 50 cm and
    (d) 200 cm from the center of the disk.

    © BrainMass Inc. brainmass.com December 15, 2022, 6:07 pm ad1c9bdddf
    https://brainmass.com/physics/electric-magnetic-fields/electric-field-due-charged-disc-125632

    Solution Preview

    ** Please see the attached file for the complete solution response **

    Fig. shows a uniformly charged disc (radius R = 0.35m), with charge density 7.9x10-3 C/m2. Let us consider an annular ring of the disc, co-centric with the disc, having radius r and width dr. Let us consider an infinitesimally small element of the ring between the radius vectors making angles θ and θ+dθ with the X axis. Area of the element = rdθdr. Charge on the element = dq = σ rdθdr where σ is the charge density.

    Electric field at point P on the axis of the disc, at a distance s from its centre, due to the element = dE = (1/4ΠЄ0)dq/l2 = (1/4ΠЄ0)σ rdθdr/l2

    From the fig. l2 = s2+r2. Hence, dE = (1/4ΠЄ0)σ rdθdr/(s2+r2) .......(1)

    The electric field vector dE at ...

    Solution Summary

    This solution provides a detailed step by step explanation of the given physics problem.

    $2.49

    ADVERTISEMENT