Question

an electron 5.00 nm to the right of a proton. a second electron is 4.00 m above the proton. what is the magnitude and direction of the net electric force on the electron force on the electron that is above the proton

Answer 1

I theink the second distance is 4 nm

Let, d1 = 5 nm

d2 = 4 nm

Electric force exerted by proton on electron, F1 = k*qe*qp/d1^2

= 9*10^9*1.6*10^-19*1.6*10^-19/(5*10^-9)^2

= 9.216*10^-12 N (towards -y axis)

so,

F1x = 0

F1y = -9.216*10^-12 N

Electric force exerted by electron on electron, F2 = k*qe*qp/(d1^2 + d2^2)

= 9*10^9*1.6*10^-19*1.6*10^-19/( (5*10^-9)^2 + (4*10^-9)^2)

= 5.62*10^-12

let theta is the angle made by F2 with +y axis,

theta = tan^-1(5/4)

= 51.34 degrees

F2x = -F2*sin(theta)

= -5.62*10^-12*sin(51.34)

= -4.39*10^-12 N

F2y = F2*cos(theta)

= 5.62*10^-12*cos(51.34)

= 3.51*10^-12 N

Fnetx = F1x + F2x

= 0 - 4.39*10^-12 N

= -4.39*10^-12 N

Fnety = F1y + F2y

= -9.216*10^-12 + 3.51*10^-12

= -5.706*10^-12 N

Fnet = sqrt(Fnetx^2 + Fnety^2)

= sqrt(4.39^2 + 5.706^2)*10^-12

= 7.2*10^-12 N <<<<<<<<<<<----------------------Answer

direction : theta = 180 - tan^-1(5.7/4.39)

= 180 - 52.4

= 127.6 degrees with +x axis<<<<<<<<<<<----------------------Answer