Recent content by moonlight13

multiply which part by that??

Can OSmeone please help me with this question.. Thank you

The dipole moment of the water molecule ({\rm H}_{2}{\rm O}) is 6.17 \times 10^{-30}\;{\rm C \cdot m}. Consider a water molecule located at the origin whose dipole moment p_vec points in the positive x direction. A chlorine ion ({\rm Cl}^{-}), of charge -1.60 \times 10^{-19}\;{\rm C}, is located...

i checked it .. that's for another problem my calculations are good for one part and now for another .. which seems strange

i gave it another shot .. and used 225000 = kQ/r^2 .. and found out r^2 .. and that worked for the second one (part B) i did the same calc but from my calculations, i am not getting the right answer .. its close but not exact ... for this part (b) i got 4.96 and the answer is 4.35 .. i...

0.0005 is from putting (1,2) in Pythagorean and then i get sqrt ((0.02)^2 + (0.01)^1) and when i did sqr of that i get 0.005 and i was given charge 28.1nC ., i figured out the E field .. now the question is if the new E vector is 225,000 in the x direction ... how do i go back .. i can't...

ok so far i got: E - kQ/r^2 = (9x10^9)(28.1x10^-9)/.0005 = 5.058x10^5 N/C .. does that look good and sensible?

well the E vector would be directed 60 deg from (0,0) .. that's what i thought .. but i am probably way off base

can someone pls guide me through this, i have 2 hours to complete this assignment and this is one of the question i am stuck with .. any help is appreciated Thank you

i am still going in circles with this .. i am no sure where am i mucking up

what you said abt the charges repelling .. this answer is right .. thank you so much once again

thank you ., thank you ... thank you..

i did the calc again and i get x = 0 y = 1.14x10^-5 .. doe sthat seem right?

so the 2 +y .. should be -y ..?

right ok .. so E = kq/r^3 = (9x10^9)(28.1x10^-9)/ (25) r= sqrt(5) ... r^3 = .0224 11290.178 N/C now .. if i take x-y-component i get .. theta = 60 deg y = 9777.58 x = 5645.1 .. does that seem right?