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(Solved): In the figure, particle 1 of charge q_(1)=4.00\times 10^(-5)C and particle 2 of charge q_(2)=3.70\ti ...



In the figure, particle 1 of charge

q_(1)=4.00\times 10^(-5)C

and particle 2 of charge

q_(2)=3.70\times 10^(-4)C

are fixed to an

x

axis, separated by a distance

d=0.400m

. Calculate their net electric field

E(x)

as a function of

x

for the following positive and negative values of

x

, taking

E

to be positive when the vector

E

points to the right and negative when

E

points to the left. What is

E(-0.1)

? Submit Answer Tries

(0)/(10)

What is

E(-0.02)

? Submit Answer Tries

(0)/(10)

What is

E(0.05)

? Tries

(0)/(10)

What is

E(0.54)

? Submit Answer Tries 0/10

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