an electron 5.00 nm to the right of a proton. a second electron is 4.00 m above

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 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 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

Explanation / Answer

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

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