(III) The dipole moment, considered as a vector, points from t...

Question

(III) The dipole moment, considered as a vector, points from the negative to the positive charge. The water molecule, Fig. 23–47, has a dipole moment $\vec{p}$ which can be considered as the vector sum of the two dipole moments $p right arrow sub 1$ and $p right arrow sub 2$ as shown. The distance between each H and the O is about 0.96 x 10^-10 m. The lines joining the center of the O atom with each H atom make an angle of 104° as shown, and the net dipole moment has been measured to be p = 6.1 x 10^-30 C m. (a) Determine the effective charge q on each H atom. (b) Determine the electric potential, far from the molecule, due to each dipole, $p right arrow sub 1$ and $p right arrow sub 2$ , and show that they sum to
V = (1 / 4π∊₀) (p cos θ / r²),
where p is the magnitude of the net dipole moment, $\vec{p} = {\vec{p}}_{1} + {\vec{p}}_{2}$ , and V is the total potential due to both $p right arrow sub 1$ and $p right arrow sub 2$ . Take V = 0 at r = ∞.

V = (1 / 4π∊₀) (p cos θ / r²),

where p is the magnitude of the net dipole moment, $\vec{p} = {\vec{p}}_{1} + {\vec{p}}_{2}$ , and V is the total potential due to both $p right arrow sub 1$ and $p right arrow sub 2$ . Take V = 0 at r = ∞.

Accepted Answer

Welcome back everyone. In this problem for a hydrogen sulfide molecule, two hydrogen sulfide bonds form an angle of 92 degrees. And the bond length between each hydrogen and sulfur atom is 1.34 multiplied by 10 to the negative 10 m. The molecule has a dipole moment of 0.97 multiplied by 10 to the negative 30th kilometers determine the expression for the electric potential V at point P which is far from the molecule caused by the dipole moments. P one and P two of the two hydrogen sulfide bonds. A hint is that we can use the relation the following relation to calculate P one and P two P one equals P two which is equal to P divided by two costs 92 degrees divided by two where P is the net dipole moment of the molecule. For our answer choices A says the electric potential is one divided by four pi epsilon, not R squared P cost. Theta B says it's one divided by two pi es two pi epsilon not R squared P costs. Theta C says it's one divided by four pi epsilon, not R squared P sine theta and D one divided by two pi epsilon knot, R squared P sine theta. Now, let's try to make sure we understand what's going on here. So we're trying to determine an expression for the electric potential V. What do we know about our electric potential as it relates to our net dipole moment? Well, recall that the electric potential V is going to be equal to one divided by four pi epsilon knot multiplied by P cost theta divided by R squared. OK. Where R is our distance as shown in the diagram, P is our net dipole moment and theta is the angle between the horizontal and or length P. So what that really means then is that on our diagram, if we draw our horizontal, OK, which is half which is at the halfway point, OK? Between our, between P one and P two, OK, then we can put our anger theta right here. OK. Now, we also know that the potential contributions from each dipole will give us the electric potential. OK. So we know that V is actually equal to V for P one, the electric potential for P one plus the electric potential for P two. Now what will be those values? OK. Well, if we think about it, our angle theta depending on P one and P two are going to be the angle from our length R and the dipole moment. So in other words, for P one, OK, or VP one, the angle we're going to consider is this angle, this angle that's marked in blue. While for P two, the angle, the value of theta that we consider is going to be this section shaded in red. Now, what are both of those angles going to be? Well, this 92 this line, sorry, as we said, it's the halfway point between P one and P two. So if it's 92 degrees between both, between both of our dipoles here, then that means that this the, the angle for both of them is going to be 46 degrees. Now, if this entire angle is 46 degrees, then that means the area shaded in blue, the angler blue is going to be 46 minus theta. Likewise, if we look at the area that's shaded in red, it's going to be over 46 degrees. OK, plus the value of theta degrees. So that means then that if we were to substitute these values to solve for electric potential, OK. That means VP one is going to be equal to one divided by four pi epsilon knot multiplied by P one costs 46 minus theta degrees divided by R squared. I should have written that in blue actually. And VP two is going to be equal to one divided by four pi epsilon knot multiplied by P two K 46 plus the degrees divided by R squared. Now, we can use that shade it in blue. No, we can use that to solve for V because now this tells us then OK, if I were to write that down here, that V is going to be equal to one divided by four pi epsilon knot, P one costs 46 minus the degrees divided by R squared plus one, divided by four pi epsilon knot multiplied by P two C 46 plus theta degrees divided by R squared. Now there are a few ways that we can simplify this. First notice that both terms have one divided by four pi epsilon knot R squared in common. Next, we also know based on the relationship given in our problem statement that P one equals P two, which is equal to P divided by two multiplied by the cosine of 92 divided by two, which is 46 degrees. So if we factor out the terms that we have in common here, then that tells us that V is going to be equal to one divided by four pi epsilon knot R squared. OK. Multiplied by P one which is P oops divided by two costs 46 degrees multiplied by the cosine of 46 minus theta degrees plus P two, which is the same expression be a divided by two cost 46 degrees multiplied by the cosine of 46 plus theta degrees. Now again, you noticed here that we have P divided by two costs 46 in common. So we can factor that out and we'll get one divided by four pi epsilon knot are squared multiplied by P divided by two cost 46. OK. Multiplied by the cosine of 46 minus the degrees plus the cosine of 46 plus the degrees. Now, we can simplify this expression by using what we know about the cosine a cosine where we're adding two angles and where we're sorry, we're adding the cosine of the difference between two angles and the sum of those two angles. In other words, since we know that the cosine of A minus B plus the cosine of A plus B is equal to two CAS A CAS B. OK. Then using that idea to simplify our expression here, which is basically the same thing where A would represent 46 and theta represents B, then we can take our electric potential even further and write it as one divided by four pi epsilon knot R squared multiplied by P divided by two cost of 46 multiplied by two cost, 46 multiplied by the cosine of theta degrees. Now, we can start to simplify, OK. Because here we can cancel two cost 46 from both the numerator and denominator and no, OK. That is going to leave us with one divided by four pi epsilon, not R squared multiplied by P cost theta. Thus, this expression will give us the electric potential at the point P which is far from the molecule. If we look back at our answer choices that tells us that a is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Dipole Moment

Electric Potential Due to a Dipole

Vector Addition of Dipole Moments

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