In the figure, particle 1 of charge q 1 = 3.50×10 -5 C and particle 2 of charge
ID: 1999856 • Letter: I
Question
In the figure, particle 1 of charge q1 = 3.50×10-5 C and particle 2 of charge q2 = 7.00×10-5 C are fixed to an x axis, separated by a distance d = 0.100 m.
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.
a) What is E(-0.010)?
-4.72×107 N/C
b) What is E(-0.040)?
-2.29×108 N/C
c) What is E(0.070)?
d) What is E(0.110)?
I got the first two right by using E= -(kq_1/0.1^2) + (kq_2/(0.1+d)^2), but can't seem to get the last two. HELP!
Explanation / Answer
eletcric field due to a point charge at distance r = kq / r^2
there is no figure, so i will assume that q1 is to the left and q2 is to the right.
there will be three case,
when location is left to the q1.
E = -[kq1/x^2] - [kq2/(x+0.1)^2]
when point is in between q1 and q2, (distance x from q1)
E = [kq1/x^2] - [kq2/(0.1-x)^2]
when point is right to the q2,
E = [kq1/(x+0.1)^2] + [kq2/x^2]
if charge location is other way around, q2 to left and q1 to the right.
when location is left to the q2.
E = -[kq2/x^2] - [kq1/(x+0.1)^2]
when point is in between q1 and q2,
E = [kq2/x^2] - [kq1/(0.1-x)^2]
when point is right to the q1,
E = [kq2/(x+0.1)^2] + [kq1/x^2]
you have not provided the figure. but these formulas will work.
us
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