Answer to Question #180292 in Physics for Rie

Question

1.what is direction of the electric field at the test point on the -x axis, 50 cm from the charge q=-5.00 nC which is located at the origin?

2.What is the electric field strength at a point 50 cm from a charge q = + 5.00 nC?

3 .A 1.65 nC charge with a mass of 1.5 x 10^-15 kg experiences an acceleration of 6.33 x 10^7 m/s2 in an electric field. What is the magnitude of the electric field?

4 Three point charges are located on the x-axis. The first charge, q1 = +10.0 µC, is at x = -1.00 m; the second charge, q2 = 20.0 µC, is at the origin; and the third charge, q3 = -30.0 µC, is located at x = 2.00 m.. What is the net force on q2?

5 what is the magnitude of the net electric field at (+1.0 m, 0) due to the three charges in number 4?

Answer 1

1) Positive x-direction.

2)

$$E=(9\cdot10^{9})\frac{5\cdot10^{-9}}{0.5^2}=180\frac{N}{C}$$

3)

$$E=\frac{ma}{q}\\E=\frac{(1.5\cdot10^{-15})(6.33\cdot10^{7})}{(1.65\cdot10^{-9})}=57.5\frac{N}{C}$$

4)

$$F=(9\cdot10^{9})\frac{(20\cdot10^{-6})(10\cdot10^{-6})}{1^2}\\+(9\cdot10^{9})\frac{(20\cdot10^{-6})(30\cdot10^{-6})}{2^2}\\F=3.15\ N$$

5)

$$F=(9\cdot10^{9})\frac{(20\cdot10^{-6})}{1^2}+(9\cdot10^{9})\frac{(10\cdot10^{-6})}{2^2}\\+(9\cdot10^{9})\frac{(30\cdot10^{-6})}{1^2}=472500\frac{N}{C}$$