Calculating Electric Potential of Ammonia Molecule: Can You Help?

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SUMMARY

The electric potential generated by an ammonia molecule, which has a dipole moment of 1.47 D, can be calculated at a distance of 52.0 nm from the dipole axis using the formula V = (1 / (4 π ε)) * (2 p / r²). Here, 'p' represents the dipole moment, and 'r' is the distance from the dipole. When an ion with a charge of q = 1.6 x 10^-19 C is brought to this point, the potential can be determined using the same formula, providing insights into the interaction between the dipole and the ion.

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  • Knowledge of the constant ε (permittivity of free space)
  • Basic algebra for manipulating equations
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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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