A 1.98 times 10 -9 C charge has coordinates x = 0, y = -2.00; a 2.73 times 10-9

Question

A 1.98 times 10 -9 C charge has coordinates x = 0, y = -2.00; a 2.73 times 10-9 C charge has coordinates x = 3.00, y = 0; and a -4.95 times 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. Determine the magnitude and direction for the electric field at the origin (measure the angle counterclockwise from the positive x-axis). magnitude direction 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). magnitude direction

Explanation / Answer

field due to 1.98*10^(-9) C charge is along +ve y-axis

magnitude=9*10^9*1.98*10^(-9)/(0.02^2)=44550 N/C

field due to 2.73*10^(-9) C charge is along -ve x-axis

magnitude=9*10^9*2.73*10^(-9)/(0.03^2)=27300 N/C

field due to -4.95 nC is along (3 ,4)/5

magnitude=9*10^9*4.95*10^(-9)/(0.03^2+0.04^2)

=17820 N/C

so x-component=17820*3/5=10692 N/C

y-component=17820*4/5=14256 N/C

so net x-component of field=10692-27300=-16608 N/C

net y-component of field=14256+44550=58806 N/C

so magnitude=61106.22 N/C

angle=105.77 degree

b)

charge of proton=1.6*10^(-19) C

so force on it=61106.22*1.6*10^(-19)=9.777*10^(-15) N

acceleration=force/mass

=5.8545*10^(12) m/s^2

angle=105.77 degree

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