SECTION 2 e-e In the above diagram, I have placed an electron (charge -e) at the

Question

SECTION 2 e-e In the above diagram, I have placed an electron (charge -e) at the origin. The grid spacing is 1 Angstrom, or 1010 m. Now place an atomic nucleus of positive charge E e (that is, times the charge of a single proton or electron) on positive x-axis, at x = E -- 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-F Angstroms? Answer: eV (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? Answer: x Angstroms

Explanation / Answer

a)

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

Q= charge on atomic nucleus = 15 x 1.6 x 10-19 C

r = distance between electron at origin and nucleus = 9.1 x 10-10 m

work done = W = kQq/r = (9 x 109) ( - 1.6 x 10-19) ( 15 x 1.6 x 10-19)/(9.1 x 10-10) = - 3.8 x 10-18 J = - 23.75 eV

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 = (15 x 1.6 x 10-19)/(d + (9.1 x 10-10))2

d = 3.2 Angstrom

so = x = - 3.2 Angstrom

c)

Velectron = Vnucleus

kq/d = k Q/(r - d)

(1.6 x 10-19)/d = (15 x 1.6 x 10-19)/(- d + (9.1 x 10-10))

d = 0.5875 A

so x = 0.5875 A

d)

r = distance of nucleus from the pint on Y-axis = sqrt(9.12 + 72) = 11.5

V = electric potential = k q/d + kQ/r = (9 x 109) (( - 1.6 x 10-19)/(9.1 x 10-10) + (15 x 1.6 x 10-19)/((11.5 x 10-10))) = 17.2 Volts

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.