give answer in 2 step with explanation at the end of each step and final answer at the end:8) An electron is released from rest at a distance of 9.00 cm from a proton. If the proton is held in place,

Question

give answer in 2 step with explanation at the end of each step and final answer at the end:8) An electron is released from rest at a distance of 9.00 cm from a proton. If the proton is held in place, how fast will the
electron move when it is 3.00 cm from the proton? (mej = 9.11 x 10-31 kg, e = 1.60 x 10-19 C, k = 1/470 = 8.99 x 109 N - m2/C2)
A) 75.0m/s B) 106 m/s €)130 m/s D) 1.06 x 103 m/s E) 4.64 x 105 m/s

9) If the electric potential in a region is given by V(x) = 6/x2, the x component of the electric field in that region is

A)-12x3. B) -6x. €)12x3, D) 12x. E) 6x.

10) The mutual inductance between two coils is 10.0 mH. The current in the first changes uniformly from 2.70 A to
5.00 Ain 0.160 s. If the second has a resistance of 0.600 Q, what is its induced current?

(A)24.0A (B)2.4A (C)240mA  (D) 24.0 mA

11) Self-inductance: At what rate would the current in a 100-mH inductor have to change to induce an emf of 1000 V
in the inductor?

A) 100 A/s B)1A/s C) 1000 A/s D) 10,000 A/s E)10A/s

12) A 45-mH ideal inductor is connected in series with a 60-Q resistor through an ideal 15-V DC power supply and an
open switch. If the switch is closed at time t = 0's, w

VIPSolutions · Accepted Answer

Let's solve each problem in **2 steps**, explaining each step, and then give the final answer.

---

### **8) Electron Released from Rest Near Proton**

**Step 1:**  
Use conservation of energy:  
$$ \frac{1}{2}mv^2 = k \frac{e^2}{r_1} - k \frac{e^2}{r_2} $$  
where $ r_1 = 9.00 $ cm, $ r_2 = 3.00 $ cm. Plug in values.

**Step 2:**  
Solve for $ v $:
$$
v = \sqrt{2k e^2 \left(\frac{1}{r_2} - \frac{1}{r_1}\right) / m_e}
$$
This gives $ v \approx 1.06 \times 10^6 $ m/s.

**Final Answer:**  
**E) 1.06 × 10^6 m/s**

---

### **9) Electric Field from Potential**

**Step 1:**  
Electric field is negative gradient of potential: $ E_x = -dV/dx $.

**Step 2:**  
Given $ V(x) = 6/x^2 $,  
$$
\frac{dV}{dx} = -12/x^3
$$
So $ E_x = 12/x^3 $.

**Final Answer:**  
**C) 12x^{-3}**

---

### **10) Induced Current from Mutual Inductance**

**Step 1:**  
Induced emf: $ \epsilon = M \frac{dI_1}{dt} $,  
$ \frac{dI_1}{dt} = \frac{5.00 - 2.70}{.160} = 14.375 $ A/s.

**Step 2:**  
Induced current: $ I_2 = \epsilon / R = (.01 \times 14.375)/.6 \approx .24 $ A = 240 mA.

**Final Answer:**  
**C) 240 mA**

---

### **11) Self-Inductance and emf**

**Step 1:**  
$ \epsilon = L \frac{dI}{dt} \implies \frac{dI}{dt} = \epsilon / L $

**Step 2:**  
$ \frac{dI}{dt} = 100 / .1 = 10,000 $ A/s.

**Final Answer:**  
**D) 10,000 A/s**

---

### **12) RL Circuit Current after 7. ms**

**Step 1:**  
Time constant: $ \tau = L/R = .045/60 = .00075 $ s = .75 ms.  
Current: $ I = \frac{V}{R}[1 - e^{-t/\tau}] $.

**Step 2:**  
$ I = 15/60 [1 - e^{-7/.75}] \approx .25 $ A = 250 mA.

**Final Answer:**  
**A) 250 mA**

---

### **13) Phase Angle in AC Circuit with Resistor**

**Step 1:**  
For a pure resistor, voltage and current are in phase.

**Step 2:**  
Phase angle is .

**Final Answer:**  
**A) **

---

### **14) RMS Voltage Value**

**Step 1:**  
$ V_{\text{rms}} = V_{\text{peak}} / \sqrt{2} $

**Step 2:**  
$ 100/\sqrt{2} \approx 71 $ V.

**Final Answer:**  
**D) 71 V**

---

### **15) Capacitors in Series**

**Step 1:**  
In series,  
- Voltages add,
- Charge same,
- Equivalent capacitance less than any single.

**Step 2:**  
All are true.

**Final Answer:**  
**D) All of the above choices are correct.**

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