# Physics

(a) Find the electric field at x = 5.00 cm in Figure 18.52(a), given that q = 1.00X10^-6;C . (b) At what position between 3.00 and 8.00 cm is the total electric field the same as that for –2q alone? (c) Can the electric field be zero anywhere between 0.00 and 8.00 cm? (d) At very large positive or negative values of x, the electric field approaches zero in both (a) and (b). In which does it most rapidly approach zero and why? (e) At what position to the right of 11.0 cm is the total electric field zero, other than at infinity? (Hint: A graphing calculator can yield considerable insight in this problem.)

Picture: q's are the particle charges, and the numbers are cm on the line.
A
-0---(+q)--5---(-2q)--10-(+q)---

B
-0-(-2q)----5---(+3q)--10----(-q)-

a.) I am having a problem with the signs, I know there should be 2 negative, and 1 positive. I also know that I should use Coulomb's law for each particle and then add them, but I'm not sure what to use for r.
b.) I'm confused about this question.
c.) I'm not sure how to approach this question, because there is no charge on 0.
d.) I understand :)
e.) Would I do the same tactic as C, but using different numbers? For example using (11+x)^2 for r in Coulomb's law.

I'm sorry for such a long question, but any help will be much appreciated! Thank you so much ahead of time!

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1. You show a charge at x = 5 but you ask for the E field there
It is UNDEFINED. 1/r^2 as r--->0 is infinite

r is simply the distance between the charge and the location where you compute E

to the left of 5, the E field due to + q at 5 is left (repels)
however the E field due to the -2q at 10 is right (attract)
perhaps at some point they add to zero
say distance d left of 5
E = -k q/d^2 + 2 k q/ (d+5)^2
can that be zero?
q/d^2 = 2q/(d^2 + 10 d + 25)
2 d^2 = d^2 + 10 d + 25
d^2 - 10 d - 25 = 0
(d-5)(d-5) = 0
d = 5
so
five to the left of +5
or at zero the field is 0

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2. d^2 -10 d - 25 = 0

d = [ 10 +/-sqrt (100+100) ]/2

= 5 +/- .5 sqrt (200)

= 5 +/- 5 sqrt 2
use +
so 5 sqrt 2 left of zero
= - 7.07

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