A 1.98 10-9 C charge has coordinates x = 0, y = 2.00; a 3.30 10-9 C charge has c

Question

A 1.98 10-9 C charge has coordinates x = 0, y = 2.00; a 3.30 10-9 C charge has coordinates x = 3.00, y = 0; and a -4.65 10-9 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.

(a) Determine the magnitude and direction for the electric field at the origin (measure the angle counterclockwise from the positive x-axis).

(b) Determine the magnitude and direction for the instantaneous acceleration of a proton placed at the origin (measure the angle counterclockwise from the positive x-axis).

Explanation / Answer

A)
let

q1 = 1.98*10^-9 C

q2 = 3.3*10^-9 C

q3 = -4.65*10^-9 C

a) E1 = k*q1/y^2

= 9*10^9*1.98*10^-9/0.02^2

= 44550 N/c

E1x = 0

E1y = 44550 N/c

E2 = k*q2/x^2

= 9*10^9*3.3*10^-9/0.03^2

= 33000 N/c

E2x = -33000 N/c

E2y = 0

E3 = k*q3/(x^2 + y^2)

= 9*10^9*4.65*10^-9/(0.03^2 + 0.04^2)

= 16740 N/c

E3x = E3*(3/5) = 16740*3/5 = 10044 N/c

E3y = E3*4/5 = 16740*4/5 = 13392 N/c

Enetx = E1x + E2x + E3x

= 0 -33000 + 10044

= -22956 N/c

Enety = E1y + E2y + E3y

= 44550 + 0 + 13392

= 57942 N/c

so, Enet = sqrt(Enetx^2 + Enety^2)

= sqrt(22956^2 + 57942^2)

= 62324 N/c <<<<<<<<--------------------Answer

Direction : theta = 180 - tan^-1(Enety/Enetx)

= 180 - tan^-1(22956/57942)

= 180-21.6

= 158.4 degrees <<<<<<<<--------------------Answer

B) Apply, F = q*Enet

m*a = q*Enet

==> a = q*Enet/m

= 2*1.6*10^-19*62324/(4*1.67*10^-27)

= 2.986*10^12 m/s^2 <<<<<<<<--------------------Answer

Directon : same as Enet. i.e 158.4 degrees <<<<<<<<--------------------Answer

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