Calculating Electric Potential of Ammonia Molecule: Can You Help?

AI Thread Summary
The discussion focuses on calculating the electric potential generated by an ammonia molecule, which has a dipole moment of 1.47 D, at a distance of 52.0 nm from its center. The formula provided for the electric potential on the dipole axis is V = (1/(4 π ε)) * (2p/r²), where p is the dipole moment. The potential needs to be determined for an ion with a charge of 1.6 x 10^-19 C brought to this point. Participants are encouraged to provide explanations and assistance in solving the problem. The request emphasizes the need for clarity in the calculations involved.
leopold123
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Could anyone help me with the following problem?
Ammonia molecule has a dipole moment of 1.47D. Calculate the potential generated by this molecule at a point on a dipole axis at a distance of 52.0 nm from the centre of the dipole. how large will the potential be if an ion of charge q=1e=1.6*10^-19 C is brought at this point? it would be really helpful if you provide any required useful explanation...thank you...waiitng eagerly for ur response...
 
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Please help me...i ll be really grateful if any1 helps me with this...
 
The electric potential at a distance 'r' on the axis of a short dipole is given by

V = \frac{1}{4 \pi \epsilon} \frac{2 p}{r^2} where p is the dipole moment which according to the problem is given to be 1.47 D.
 

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