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quicknote
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Can somebody please check over my solution. I cannot figure out what's wrong. My final answer is not right. I think I may have an error with the signs of the point charges or may have a dumb calculation error.
Thank youFour point charges, fixed in place, form a square with side length d.
The particle with charge q is now released and given a quick push; as a result, it acquires speed v. Eventually, this particle ends up at the center of the original square and is momentarily at rest. If the mass of this particle is m, what was its initial speed v?
See picture here: http://ca.geocities.com/canbball/index.html
By using energy conservation I get:
[tex]k_{i} = U_{f} -U_{i} [/tex]
[tex] \frac{\ 1}{2} *mv^2 =k[( \frac{\ -3q^2}{\frac{\ d}{\sqrt{2}}} + \frac{\ 5q^2}{\frac{\ d}{\sqrt{2}}} + ( \frac{\ 2q^2}{\frac{\ d}{\sqrt{2}}}) - (( \frac{\ -3q^2}{d} + ( \frac{\ 2q^2}{d} + ( \frac{\ 5q^2}{d})] [/tex]
after simplifying, I finally get:
[tex] v= \sqrt {\frac{\ 3.31kq^2}{md}} [/tex]ETA: Sorry, I'm not sure how to fix the tex...
Thank youFour point charges, fixed in place, form a square with side length d.
The particle with charge q is now released and given a quick push; as a result, it acquires speed v. Eventually, this particle ends up at the center of the original square and is momentarily at rest. If the mass of this particle is m, what was its initial speed v?
See picture here: http://ca.geocities.com/canbball/index.html
By using energy conservation I get:
[tex]k_{i} = U_{f} -U_{i} [/tex]
[tex] \frac{\ 1}{2} *mv^2 =k[( \frac{\ -3q^2}{\frac{\ d}{\sqrt{2}}} + \frac{\ 5q^2}{\frac{\ d}{\sqrt{2}}} + ( \frac{\ 2q^2}{\frac{\ d}{\sqrt{2}}}) - (( \frac{\ -3q^2}{d} + ( \frac{\ 2q^2}{d} + ( \frac{\ 5q^2}{d})] [/tex]
after simplifying, I finally get:
[tex] v= \sqrt {\frac{\ 3.31kq^2}{md}} [/tex]ETA: Sorry, I'm not sure how to fix the tex...
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You're right...one of the them is [tex] \sqrt{2} [/tex] away.