The figure shows an uneven arrangement of electrons (e) and protons (p) on a cir

Question

The figure shows an uneven arrangement of electrons (e) and protons (p) on a circular arc of radius r =1.99 cm, with angles teta_1 7degree, teta_2 = 50degree,teta_3=30degree, teta_4 = 20degree. What are the magnitude and direction (relative to the positive direction of the x axis) of the net electric field produced at the center of the arc? In the figure below, eight charged particles form a square array; charge q = -e and distance d = 4.0 cm. What are the magnitude and direction of the net electric field at the center?

Explanation / Answer

Horizontal component of electric field

= 8.98 *109 * 1.6 *10-19/0.01992 - 8.98 *109 * 1.6 *10-19*cos7/0.01992 - 8.98 *109 * 1.6 *10-19*cos57/0.01992 - 8.98 *109 * 1.6 *10-19*cos50/0.01992 + 8.98 *109 * 1.6 *10-19*cos20/0.01992

=   3.628 * 10-6 - 3.601 * 10-6 - 1.976 * 10-6 - 2.332 *10-6 + 3.409 * 10-6

= - 0.872 * 10-6 N/C

Vertical component of electric field

= - 8.98 *109 * 1.6 *10-19*sin7/0.01992 - 8.98 *109 * 1.6 *10-19*sin57/0.01992 + 8.98 *109 *1.6*10-19*sin50/0.01992 - 8.98 *109 * 1.6 *10-19*sin20/0.01992

=    - 0.442 * 10-6 - 3.0428 * 10-6 + 2.779 *10-6 - 1.2409* 10-6

=      - 1.9467 * 10-6   N/C

=> Magnitude of Electric field = 2.133 * 10-6 N/C

=> Direction = 180 + tan-1(1.9467/0.872)

                        = 245.87   degrees            relative to + ve X axis (anticlockwise)

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