Electrostatics Net electric field net electric force net electric potential Two

Question

Electrostatics

Net electric field

net electric force

net electric potential

Two point charges (q_1 and q_2) are located on the x-axis respectively at x = 0 and x = d, as shown below. Let q_1 = +90 mu C, q_2 = -60 mu C, and d = 8.0 cm. (a) Find the magnitude and direction of the net electric field on the y-axis at y = -6.0 cm. (Specify the direction as an angle measured counterclockwise from the +x axis.) (b) Suppose a -25 mu C test charge is placed on the y-axis at y = -6.0 cm. Determine the magnitude and direction of the net electric force on the test charge. Again, specify the direction as an angle measured counterclockwise from the +x axis. (c) The net electric potential is zero at a certain point on the x-axis to the right of q_2. Determine the x coordinate of this point.

Explanation / Answer

a) Ex = Summation kq/r^2

= 9e9*60e-6/(0.06^2+0.08^2) *0.08/sqrt(0.06^2+0.08^2)

= 4.32*10^7 N/C

Ey = - 9e9*90e-6/0.06^2 + 9e9*60e-6/(0.06^2+0.08^2) *0.06/sqrt(0.06^2+0.08^2)

= -19.26*10^7 N/C

Enet = sqrt(19.26^2+4.32^2) *10^7

Enet = 1.974*10^8 N/C

direction = 270 degree + arctan(4.32/19.26)

= 282.6 degree above x axis.

b) F= qE = -25e-6*1.974e8 = 4935 N

direction = 282.6-180 degree

= 102.6 degree above x axis CCW

c) V = 0

kq1/r1+kq2/r2 = 0

90/x = 60/(x-8)

3x -24 = 2x

x = 24 cm Answer

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