In each of the figures 1(a), 1(b), and 1(c) a particle with a positive charge of
ID: 1527246 • Letter: I
Question
In each of the figures 1(a), 1(b), and 1(c) a particle with a positive charge of magnitude q is moved a distance d in an electric field E. The magnitude of the charge, the distance d. and the strength of the electric field E are the same for all three figures. In figure 1a the path along which the particle is moved is parallel to the field, in figure 1(b) it is perpendicular to the field, and in figure 1(c) it makes a 45_ angle. In which figure is the work done by the field greatest? In which figure is it least? Explain your answers. If q = 5 times 10-9 C. E = 500N/C, and d = 0.02 m, calculate the amount of work done by the field on the particle in each of the figures.Explanation / Answer
Solution:
In figure A, the positive gharge moves in the same direction as electric field, so work done will be qXEXd.
In figure B, the charge q moves in the perpendicular durection of electric field, so work done by electric field will be zero, as in that displacement direction, there will be no force due to electric field.
In figureC. the work done by electric field = qXEXd cos45 =qXEXd/1.41 as displacement in the direction of electric field is d cos45.
SO, work done is figure A will be highest and work done in figure B will be lowest.
Work done in Case A = 5X10^-9X500X0.02 = 5X10^-8 joule
Work done in Case C = 5X10^-8/1.4 = 3.57X10^-8 joule
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.