Question 1 The charges in the figure above both have a magnitude of 4.00 nC, and

Question

Question 1 The charges in the figure above both have a magnitude of 4.00 nC, and the distance d is 9.00 cm. What is the magnitude of the electric field at the dot? 8.88E+ 03 N/C 4.44E-03 N/C 3.14E 03 N/C 6.28E+03 N/C Question 2 In the figure above, charge A is -3.50 nC, charge B is 7.00 nC, and charge C is 3.50 nC. If x3.10 cm and y 6.20 cm, what is the electric fleld at the dot? o4.20x104 N/C, 19.4 counter-clockwise from the positive x axis 4.20x104 N/C, 38.9clockwise from the positive xaxis 3.16x104 N/C, 19.4° clockwise from the positive x-axis 3.16x10 N/C, 38.9 clockwise from the positive xaxis Question3 A charged 3.50 g glass bead hangs suspended in the air by an electric field. If the bead was charged by the removal of 4.40x1010 electrons, what must the electric field magnitude be? 1.541011 N/c 4.97 108 N/C O 4.87x10 N/C O 7.80x10 N/C 13

Explanation / Answer

Q3.

Q = n*e

e = 1.6*10^-19

n = no. of electrons removed

Q = 4.4*10^10*1.6*10^-19

Q = 7.04*10^-9 C

Now using force balance on bead

Fe = Fg

Fe = electrostatic force, which is given by

Fe = Q*E

Fg = gravitational force = m*g

So,

Q*E = m*g

E = m*g/Q

Using known values:

E = 3.5*10^-3*9.81/(7.04*10^-9)

E = 4.877*10^6 N/C

Correct option is C.

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

---

---

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