arolina State University-P 255 - Spring18-CASH Activities and Due Dates HW5: Ele

Question

arolina State University-P 255 - Spring18-CASH Activities and Due Dates HW5: Electric Fields and Forces 2/16/2018 11:59 PM Gradeboa 0/100 Pint Calaitor-Perode Table Question 6 of 10 Sapling Learning A nucleus of the boron-11 isotope consists boron-11, whose mass is 1.83 × 1 ot tive protons and six neutrons. A particular ionized atom of kg, lacks 3 electrons from its neutral state. Find the magnitude and direction of the electric field that will levitate this ion, exactly balancing its weight. Take g = 9.81 mr Number N/C Direction: O Upward Cannot be determined O Downward Horizontal PreviusCock Answer O Next E Hint

Explanation / Answer

The force from an electric field is F = qE   Since the ion is missing four electrons, it has a poisitve charge of

(3)(1.6 x 10^-19) = 4.8 x 10^-19 C

the electric force must match the weight of the ion

so qE = mg

E = mg/q

E = (1.83 x 10^-26)(9.81) / (4.8 x 10^-19)

E = 3.94 X 10^-7 N/C

Since the charge of the is positive, the Field will have to point upwards.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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