iPad 4:41 PM 54% c MyCr Blackboard Learn ×\' GPll-Chapter19 × \' Practice Anothe
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iPad 4:41 PM 54% c MyCr Blackboard Learn ×' GPll-Chapter19 × ' Practice Another × ' Practice Another x a Two Point Charge Kwww.webassign.net/web/Student/Assignment-Responses/last?dep 13329158 7. 0/10 pointsI Previous Answers Walker4 19.P.045 My Notes Two point charges lie on the x axis. A charge of-7.5 pC is at the origin, and a charge of +3.5 C is at x 10.0 cm. a) What is the net electric field at x2.0 cm? 14.8e+07 X The response you submitted has the wrong sign.x N/C (b) what is the net electric field at x = +2.0 cm? 2.1e+07 X The response you submitted has the wrong sign. x N/C 8. 10/10 points | Previous Answers Walker4 19.P053. My Notes The electric field lines surrounding three charges are shown in the figure below. The center charge is q2 =-19.3 pC. E field linesExplanation / Answer
a) at x = -2 cm
The electric field due to charge at origin,
E1 = kq/r^2 x^ = 9*10^9*7.5*10^-6/(2*10^-2)^2 = 1.6875 *10^8 N/C x^
The electric field due to charge at x = 10cm,
E2 = kq/r^2 -x^ = 9*10^9*3.5*10^-6/(12*10^-2)^2 = 2.187500*10^6 -x^
Net electric field E = 166.5625*10^6 = 1.67*10^8 N/C x^.
b) At x = +2 cm
The electric field due to charge at origin,
E1 = kq/r^2 -x^ = 9*10^9*7.5*10^-6/(2*10^-2)^2 = 1.6875 *10^8 N/C -x^
The electric field due to charge at x = 10cm,
E2 = kq/r^2 -x^ = 9*10^9*3.5*10^-6/(12*10^-2)^2 = 2.187500*10^6 -x^
Net electric field E = 170.9375*10^6 = 1.71*10^8 N/C -x^.
The net electric field E =
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