Homework Answers
Q.20.24)
Enet = 2*E*cos(45 deg)
NOTE: The vertical components of the electric fields will cancel out and the horizontal components will add up
So, Enet = 2*(k*q/r^2)*cos(45 deg)
= 2*(9*10^9*1*10^-9/(0.05^2 + 0.05^2))*cos(45 deg)
= 2545.6 N/C
b)
direction = 0 deg (the net electric field will be horizontal)
2)
Enet = 4*(k*q/r^2)*cos(45 deg)
= 4* (9*10^9*2*10^-9/(0.02^2 + 0.02^2))*cos(45 deg)
= 63639 N/C
So, force = Enet*q = 63639*(3.5*10^-9)
= 2.23*10^-4 N <----------- your answer's unit is wrong i guess. It should be in N
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Part A.) The electric flux through a spherical surface is
4.0\times 10^4~\text{N}\cdot\text{m}^2/\text{C}4.0×10 4 N⋅m 2
/C. What is the net charge enclosed by the surface?
Part B.) The electric field 10.0 cm from the surface of a copper
ball of radius 5.0 cm is directed toward the ball’s center and has
magnitude 9.0*10^2N/C. How much charge is on the surface of the
ball?
Select the correct answer
a.) 2.2*10^{-10} . b.)-1.25*10^{-10} . c.)-1.0* 10^{-9} .
d.)-4.5* 10^{-10) ....
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