Question
Part 1 of 2) A charge of 5.16 µC is at the origin and asecond charge of 2.60 µC is on the positivex-axis 1.01 m from the origin.Find the magnitude of the electric eld ata point P on the y-axis 0.581 m from theorigin. The value of the Coulomb constant is8.99 × 109N · m2/C2.Answer in units of N/C.
Part 2 of 2) Determine the direction of this electric eld(as an angle between 180and 180 measured from the positive x-axis, with counterclockwise positive).Answer in units of.
Part 1 of 2)A 2.3 µC point charge is on the x-axis atx = 2.8 m , and a 4 µC point charge is on thex-axis at x = 1.3 m .Find the magnitude of the net electriceld at the point on the y-axis where y =4.9 m . The value of the Coulomb constant is8.98755 × 109N · m2/C2.Answer in units of N/C.
Part 2 of 2) Determine the direction of this electric eld(as an angle between 180and 180 measured from the positive x-axis, with counterclockwise positive.)Answer in units of.
Explanation / Answer
Given 1) q1 = 5.16 C is at (0 , 0) q2 = - 2.60 C is at (1.01 m , 0) Electric field at point P (0 , 0.581m ) due to q1 and q2 is E = k q1 / r1^2 (+j) + k q2 / r2^2 [ cos (i) + sin (-j) ] ---1 here r2 = (0.581 m)^2 + (1.01 m)^2 = 1.165 m And = sin-1(0.581 /1.165) = 300 Taking eq 1, E = (8.99 *10^9 N-m^2/C^2) (5.16 *10^-6 C) / (0.581m)^2 (j) + (8.99 *10^9 N-m^2/C^2) (2.60 *10^-6 C) / (1.165 m)^2 ( cos30 (i) + sin 30 (-j)) E = 1.37 *10^5 (j ) + 1.49 *10^4 (i) + 8.61 *10^3 (-j) E = 1.49 *10^4 (i) + 12.8 *10^4 (j) Magnitude of electric field is E = ( 1.49 *10^4)^2 + (12.8 *10^4)^2 = 1.28 *10^5 N/ C -------------------------------------------------------------------------------------------- 2) Direction of electric field is = tan-1(12.8 / 1.49) = 83.36 0 counterclock wise from x axis *post only one question for single post* Taking eq 1, E = (8.99 *10^9 N-m^2/C^2) (5.16 *10^-6 C) / (0.581m)^2 (j) + (8.99 *10^9 N-m^2/C^2) (2.60 *10^-6 C) / (1.165 m)^2 ( cos30 (i) + sin 30 (-j)) E = 1.37 *10^5 (j ) + 1.49 *10^4 (i) + 8.61 *10^3 (-j) E = 1.49 *10^4 (i) + 12.8 *10^4 (j) Magnitude of electric field is E = ( 1.49 *10^4)^2 + (12.8 *10^4)^2 = 1.28 *10^5 N/ C -------------------------------------------------------------------------------------------- 2) Direction of electric field is = tan-1(12.8 / 1.49) = 83.36 0 counterclock wise from x axis *post only one question for single post*