Answer to Question #166061 in Physics for Audrey

Question

1. Two unknown charges attract each other. The first charge has twice greater magnitude than the second charge. If the force acting on the charges is 20^-12 N and the distance between them is 120 nm, what is the charge of the second charge? What is the force at 200 nm?
2. A 7.0 x 10^-9 C point charge is at the center of a 5.0-cm radius sphere. What is the electric flux through the sphere surface?
3. A charge has a magnitude of 27 picocoulomb (1 pC=10^-12 C) and is in a uniform electric field E. It is released from rest and moves. After it has moved for about 18 pm, it is found that its kinetic energy is 150 nanojoule (1 nJ=10^-9 J). (a) What is the electric potential of the starting point with respect to the endpoint?
4. A small sphere with a charge of -0.60 microcoulomb is placed in a uniform electric field of magnitude 1.2x 10⁶ N/C pointing to the west. What is the magnitude and direction of the electric field?
5. A copper wire (p=1.724 x 10^-8 ohm m) has a length of 30 m and a diameter of 2 mm. If the power on the wire is 2 x 10^-4W, what is the current flowing on the wire?
6. A 2.5-ampere current flows through a silver wire with a radius of 23 mm. The wire is powered by a 220 V electromotive force. (a) What is the length of the wire? (b) What is the power dissipated by the wire?
7. An electrician wishes to cut a copper wire (p=1.724 x 10^-8 ohm m) that has more than 10 ohm resistance. The wire has a radius of 0.725 mm. Approximately what length of the wire has a resistance equal to the maximum 10 ohm?

Answer 1

1)

$$2\cdot10^{-12}=9\cdot10^9\frac{2q^2}{(120\cdot10^{-9})^2}\\q=1.26\cdot10^{-18}\ C$$

$$F'=\frac{120^2}{200^2}2\cdot10^{-12}=7.2\cdot10^{-13}N$$

2)

$$\Phi=\frac{7\cdot10^{-9}}{8.85\cdot10^{-12}}=790\ Vm$$

3)

$$V=\frac{150\cdot10^{-9}}{27\cdot10^{-12}}=5556\ V$$

4)

$$F=(0.6\cdot10^{-6})(1.2\cdot10^{6})=0.72\ N$$

Eastwards.

5)

$$2 \cdot 10^{-4}=1.724 \cdot 10^{-8}\frac{30}{\pi (1\cdot 10^{-3})^2}I^2\\I=0.035\ A$$

6) a)

$$\frac{220}{2.5}=1.59 \cdot 10^{-8}\frac{l}{\pi (23\cdot 10^{-3})^2}\\l=9.2\cdot 10^{6}\ m$$

b)

$$P=220(2.5)=550\ W$$

7)

$$R=\rho\frac{l}{\pi r^2}\\10=1.724 \cdot 10^{-8}\frac{l}{\pi (0.725\cdot 10^{-3})^2}\\l=959\ m$$