Question

A 2.18e-9 C charge has coordinates x = 0, y = −2.00; a 2.76e-9 C charge has coordinates x = 3.00, y = 0; and a -4.95e-9 C charge has coordinates x = 3.00, y = 4.00, where all distances are in cm. Determine magnitude and direction for the electric field at the origin and the instantaneous acceleration of a proton placed at the origin.

Answer 1

given

charges are

q1 = 2.18*10^-9 C , at (x1,y1) = (0,-2.0 cm)

q2 = 2.76*10^-9 C , at (x1,y1) = (3.0,0 cm)

q3 = -4.95*10^-9 C , at (x1,y1) = ( 3.0,4.0 cm)

the reference point is at origin

we know that the electric field due to a charge at a distance r is E = kq/r^2

by symmetry the directions of the fields due charges are  

q1 , is along +y direction  

q2, is along -x direction

q3 is making an angle theta = arc tan (4/3) with +x axis that is theta = 53.13 degrees

so the magnitudes of the fields due to charges at origin are

E1 = k*q1/r1^2 N/C = (9*10^9*2.18*10^-9/(-0.02)^2) N/C = 49050 N/C

E2 = k*q2/r2^2 N/C = (9*10^9*2.76*10^-9/(0.03)^2) N/C = 27600 N/C

E3 = k*q3/r3^2 N/C = (9*10^9*4.95*10^-9/(0.05)^2) N/C = 17820 N/C

writing the field in terms of components

E1 = E1 cos theta i + E1 sin theta j = 49050 cos90 i + 49050 sin 90 j = 0 N/C i + 49050 N/C j

E2 = E2 cos theta i + E2 sin theta j = 27600 cos(180) i + 27600 sin (180) j = -27600 N/C i + 0 N/C j

E3 = E3 cos theta i + E3 sin theta j = 17820 cos(53.13) i + 17820 sin (53.13) j = 4692 N/C i + 14256 N/C j

the net field E = Ex i + Ey j

Ex = (0 -27600 +4692 )N/C i = -22908 N/C i

Ey = (49050 +0 +14256)N/C j = 63306 N/C j

the magnitude is  

E = sqrt(Ex^2+Ey^2) = sqrt((-22908)^2+(63306)^2) N/C = 67323.295 N/C

the direction is theta = arc tan (Ey/Ex) = arc tan((63306)/(-22908))= -70.10 degrees= 180-70.10 degrees = 109.90 degrees with +x axis

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if proton is at origin, so that the like charges reppel and unlike charges attracts

so the force direction is also as same as the field direction here

F1 = E*q N ==> F1 = 49050 N/C*1.6*10^-19 C = 78480*10^-19 N ==>F1 = 0 N i +78480*10^-19 N j

F2 = 27600 N/C*1.6*10^-19 C = 44160*10^-19 N = 44160*10^-19 N==>F2 = -44160*10^-19 N i + 0 N j

F3 = 17820 N/C*1.6*10^-19 C = 28512*10^-19 N = 28512*10^-19 N

now the force F3 = 28512*10^-19 cos(53.13) i + 28512*10^-19 sin (53.13) j = 17107.24*10^-19 N/C i + 22809.57*10^-19 N/C j

Net force is F = (0 -44160*10^-19 +17107.24*10^-19) N i + (78480*10^-19+0+22809.57*10^-19) N j

F = (-2.705)*10^-15 N i + 1.013*10^-14 N j

manitude is F = sqrt(((-2.705)*10^-15)^2+(1.013*10^-14)^2) N = 1.0485*10^-14 N

F= ma

a = F/m

mass of the proton is m = 1.6726*10^-27 kg

a = (1.0485*10^-14)N /(1.6726*10^-27) kg

a = 6.267*10^12 m/s2

the instantaneoud acceleration of the proton is a = 6.267*10^12 m/s2