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Chapter 4: Q56E (page 717)

The dipole moment of the water molecule (H2O) is6.17脳10鈭�30C/m. Consider a water molecule located at the origin whose dipole moment p points in the +x direction. A chlorine ion (Cl-), of charge鈭�1.60脳10鈭�19C, is located atx=3.00脳10鈭�9m. Find the magnitude and direction of the electric force that the water molecule exerts on the chlorine ion. Is this force attractive or repulsive? Assume that x 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 derived in Example 21.14 can be used.

Short Answer

Expert verified

The direction of the force is in negative x direction and the magnitude is 6.58脳10鈭�13N

Step by step solution

01

Electric field due to the dipole

Calculate the force that is exerted by the electric dipole by

F鈫�=qE鈫�

where q= charge and E=electric field applied

Electric field applied by the electric dipole in the x direction is:

Ex=12蟺蔚opx3

where 14蟺蔚0=9.0脳109N鈰�m2/C2

02

Force exerted due to the electric field

From the above equations calculate F:

F=12蟺蔚0qpx3=18.0脳109N鈰�m2/C2鈭�1.6脳10鈭�19C6.17脳10鈭�30Cm3脳10鈭�9m3=鈭�6.58脳10鈭�13N

Since the value of force F is negative the direction is in negative x direction.

Therefore, the direction of the force is in negative x direction and the magnitude is 鈭�6.58脳10鈭�13N.

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