Question

I need help with the second half of my problem. I believe this would fit in the category of calculusII. I'm going to take a little time her to type the full question, i only need help with part b.

Answer 1

a. In this problem we will examine an "electric dipole". Many system can be viewed as consisting of equal and opposite electric charges separated by some distance (e.g. the salt molecule NaCl). thus, it is important to understand their cumulative electrical effect on their surroundings. using a=0.1 m and d=1 m, determine the net electric force on q. Then repeat this but use d=2m. by what factor did the net electric force change? Explain this result. Note that the distance d extend from q to the point halfways between the other two charges.

Answer 2

here is the work and answer for a:
\[F_{Q_{1}q}=\frac{kqQx}{1.1025}+ F_{Q_{2}q}=\frac{kqQx}{0.9025}=0.2kqQxN\]
\[F_{Q_{1}q}=\frac{kqQx}{4.2025}+ F_{Q_{2}q}=\frac{kqQx}{3.8025}=0.02kqQxN\]
changes by a facto of 10 because the further away the weaker the force.

Answer 3

Those should be subtracted my mistake. and reversed

Answer 4

ok b.
now used a binomial expansion to expand the electric force from each charge in the dipole in terms of a/(2d). show that after "adding" the electric forces due to each charge that the first remaining terms depends upon 1/d^3. here is the binomial expansion if |x|<1.

Answer 5

\[(1+x)^{m}=1+\sum_{\infty}^{n=1}\frac{m(m-1)(m-2)...(m-n+1)x^{n}}{n!}\]

Answer 6

i put the infinity and the n=1 on the wrong ends =/ sorry

Answer 7

Are you sure the upper limit is \(\infty\)?

Answer 8

umm let me check really quick

Answer 9

I guess it doesn't hurt because the terms will be 0 for \(n\gt m\)

Answer 10

yeah upper limit is infinity.