Trig question in a physics problem about potential change from an electric field

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Trig question in a physics problem about potential change from an electric field

The pic below part b is wrong I redid the problem correctly however the trig part at the end i dont understand how to get cot 45 and what that represents. thanks!

693 ELD 24-2 EQUIPOTENTIAL SURFACES AN ELECTR Problem 24.02 Finding the potential change from the electric field shows two p field E. The pol d of the test charge is perpendicular to E. Thus, th angle obetween E and d s is 90 and the dou product E.ds and f n a uniform elec- on the same electric field line (no shown) and are sepa on 24-18 then tells us that points i and c are at the ed by a distance d. Find the poten s 0. Equi difference v 0. Ah, we should have seen this by moving a positive test charge qu from same potential: V Vi along the path shown, which is parallel to the field coming. The points are on the same equipotential surface direction which is perpendicular to the electric field line For line cf we have 45 and, from Eq-24-18 KEY IDEA E(cos 450) ds We can find the potential difference between any two points in an electric field by integrating E.d3 along a path con necting those two points according to Eq.24-18 E(cos, 4 ds. Calculations: We have actually already done the calculation for such a path his equation is just the length of he direction of an electric field line The integral ength is d/cos 450, Thus, uniform field when we derived Eq 24-21. With slight changes in from Fig. 24-8b, tha notation, Eq. 24-21, gives us Ed. (Answer E(cos 450) cos 45 Ed. (Answer (b) Now find the potential This is the same result we obtained in (a), as it must be difference V V by moving the the potential difference between two points does not de positive test charge q from i to falong the path icf shown in pend on the path connecting them. Moral: When you Fig.24 want to find the potential difference between two points Calculations: The Key Idea of (a) app by moving a test charge between them, you can save ere too, except and work by choosing a path that simplifies the use of now we move the test charge along a path consists of two lines: ic and cf At all points along line ic, he displace- Eq. 24-18. The electric field points from The field is perpendicular to this ic path, higher potential to lower potential so there is no change in the potential. Higher potential The field has a component along this cf path, so there is a change in the potential. Lower potential Figure 24 B (a)A test charge go moves in a straight line from point ito point along the direction of a uniform external electric field. (b) Charge qo moves along path icf in the same electric field PLUS Additional examples, video, and practice available at WileyPLUS

Explanation / Answer

part b:

potential difference=-(integration of dot product of E and ds)

let horizontal be +ve x axis and vertical be +ve y axis.

so electric field is along -ve y axis.

in vector notation, electric field=-E j

in the triangle icf,

length of if=d

angle icf=45 degrees

then length of cf=length of if/sin(45)=1.4142*d

now path cf lies 45 degree below -ve x axis.

so in vector notation, path cf=1.4142*d*(-cos(45) i- sin(45) j)

=-d i -d j

so potential change=-(dot product of E and path cf)

=-(dot product of (-E j) and (-d i -d j)

=-(-E*(-d))

=-E.d

which is same as the answer obtained in part a.

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