A point charge -3.61nC is at the point x = 0.60 m, y = 0.80 m , and a second poi

Question

A point charge -3.61nC is at the point x = 0.60 m,y = 0.80 m , and a second point charge 6.48nC is at the point x = 0.60 m , y = 0.

A. What is the magnitude of the electric field set up at the origin from q1?

B. What is the magnitude of the electric field set up at the origin from q2?

C. What is the x component of the net electric field at the origin?

D. What is the y component of the net electric field at the origin?

E. Calculate the magnitude of the net electric field at the origin due to these two point charges.

F. Calculate the direction of the net electric field at the origin due to these two point charges. ( counter-clockwise from + x-axis)

Explanation / Answer

Distance of q1 from origin=r1= sq rt(0.6)^2 + (0.8)^2=1 m

Distance of q2 from origin=r2=0.6 m

Electric field due to q1 =E1=kq1/r1^2

Electric field due to q1 =E1=9*10^9*3.61*10^-9/1^2

Electric field due to q1 =E1=32.41 N/C at angleO with x axis,where tanO=53 degree

Component of E1 along x axis =32.41cos53=19.55 N/C (i)

Component of E1 along y axis =32.41sin53=25.88 N/C (j)_______________________________
Electric field due to q2 =E2=kq2/r2^2

Electric field due to q2 =E2=9*10^9*6.48*10^-9/0.6^2

Electric field due to q2 =E2=162 N/C along (-) x axis

In vector notation,E2 =162N/C(-i)
_____________________________

Net field = E = E1+ E2

E = 19.55 N/C (i) + 25.88 N/C (j) + 162N/C(-i)

E = 142.45 N/C (-i) + 25.88 N/C(j)

Magnitude of resultant = E =sq rt 17316 =144.78 N/C

Angle O which Emakes with negative x axis is given by tanO= 25.88/142.45=0.18

Angle O =10.29 degree clockwise with negative x axis or 190.29 degree clockwise with positive x axis

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