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moonlight13
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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 at x=3.00 \times 10^{-9} meters. Assume that this x value is much larger than the separation d between the charges in the dipole, so that the approximate expression for the electric field along the dipole axis can be used.
Find the magnitude of the electric force, ignoring the sign, that the water molecule exerts on the chlorine ion.
attempt to question:
the water has an electric field to the right and the chlorine ion has it to the left ..
i calculated the E for water and Cl- seperately
E (H2O):
E = 2Kp/r^3 = 2*(9x10^9)*(6.17x10^-30)/(3.00x10^-9)^3 = 4.11 x 10^6 N/C
E (Cl-):
E = KQ/r^2 = K(1.60x10^-19)/(3.00x10^-9)^2 = 1.6 x 10^8 N/C
now to find the force do I find the E(net) = 1.6 x 10^8 - 4.11 x 10^6 = 1.56 x 10^8 N/C
then times by charge? .. so E(net) x (1.6 x 10^-19) ..? = 2.49 x 10^-11 N...
Am I doing it right..??
Thank
Find the magnitude of the electric force, ignoring the sign, that the water molecule exerts on the chlorine ion.
attempt to question:
the water has an electric field to the right and the chlorine ion has it to the left ..
i calculated the E for water and Cl- seperately
E (H2O):
E = 2Kp/r^3 = 2*(9x10^9)*(6.17x10^-30)/(3.00x10^-9)^3 = 4.11 x 10^6 N/C
E (Cl-):
E = KQ/r^2 = K(1.60x10^-19)/(3.00x10^-9)^2 = 1.6 x 10^8 N/C
now to find the force do I find the E(net) = 1.6 x 10^8 - 4.11 x 10^6 = 1.56 x 10^8 N/C
then times by charge? .. so E(net) x (1.6 x 10^-19) ..? = 2.49 x 10^-11 N...
Am I doing it right..??
Thank
