In the above diagram, I have placed an electron (charge -e) at the origin. The g

Question

In the above diagram, I have placed an electron (charge -e) at the origin. The grid spacing is 1 Angstrom, or 10^-10m. Now place an atomic nucleus of positive charge 8e (that is, 8 times the charge of a single proton or electron) on positive x-axis, at x= 3.4 angstroms. Now answer the following questions: a)How much work did it take you to bring this nucleus in from very far away and place it on the positive x-axis at x=3.4 angstroms? b) There will be a place on the x-axis where the total electric field due to the two charges is zero. At what value of x will this occur? C)Looking along the x-axis, at what value of x will the total electrostatic potential V be zero? Answer on the line given here, and also sketch on the diagram above the approximate shape of the V=0 equipotential surface. d)What is the value of the electrostatic potential V at a point on the positive y-axis, at y=6.1 angstroms e) How much work would it take to bring in another atomic nucleus of positive charge 4 e from very far away and place it at this point on the positive y-axis at y=6.1 Angstroms?

Explanation / Answer

a)

q = charge on electron = - 1.6 x 10-19 C

Q= charge on atomic nucleus = 8 x 1.6 x 10-19 C

r = distance between electron at origin and nucleus = 3.4 x 10-10 m

work done = W = kQq/r = (9 x 109) ( - 1.6 x 10-19) ( 8 x 1.6 x 10-19)/(3.4 x 10-10) = - 5.42 x 10-18 J

b)

let the point be distance "d" to the left of origin

Eelectron = Enucleus

kq/d2 = k Q/(r + d)2

(1.6 x 10-19)/d2 = (8 x 1.6 x 10-19)/(d + (3.4 x 10-10))2

d = 1.86 Angstrom

so x = - 1.86

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