Give step-by-step solution with explanation and final answer:PhET Simulation: Charges and Fields. Ek
Part 1. The electric potential (voltage) at a specific location is equal to the potential energy p

Question

Give step-by-step solution with explanation and final answer:PhET Simulation: Charges and Fields. Ek
Part 1. The electric potential (voltage) at a specific location is equal to the potential energy per unit
charge a charged object would have if it were at that location. If the zero paint of the voltage is at infinity, 3
the numerical value of the voltage s equal to the numerical value of work Gone fo bring in @ unit charge
from infinity to that location
Select Show numbers and grid in the green menu, and drag one positive charge to the middle of the
screon. right on top of two intersecting bold grid fines.
Using the voltage meter, ou should find that 1m away fom the charge, the voltage is OV.
1) Whats the voltage 2 m away from the charge?
ANSWER: LI.5\|
2) Whatis the yotage 3m away from the charge?
ANSWER: 3.0)
Part 2. Another way to study voltage and its relationship to electric fied is by producing equipotental
lines. Just like every paint on a contour line has the same elevation in a topographical map, every point
on an equipotenial ine has the same voltage.
Click plot on the voltage tool to produce an equipotential fine. Produce many equipotent lines by
clicking

VIPSolutions · Accepted Answer

Let's solve each part step by step:

---

### **Part 1**

Given:  
- At 1 m from the charge, the voltage is **9 V**.

The electric potential $ V $ from a point charge falls off as $ V \propto \frac{1}{r} $.

#### 1. What is the voltage 2 m away from the charge?

Since voltage is inversely proportional to distance:
$$
V_2 = V_1 \times \frac{r_1}{r_2} = 9 \text{ V} \times \frac{1}{2} = 4.5 \text{ V}
$$

**ANSWER:** $ 4.5 $ V

---

#### 2. What is the voltage 3 m away from the charge?

$$
V_3 = V_1 \times \frac{r_1}{r_3} = 9 \text{ V} \times \frac{1}{3} = 3 \text{ V}
$$

**ANSWER:** $ 3.0 $ V

---

### **Part 2**

#### 3. Which statement is true?
Let's analyze the options:

- **a)** At any point, the electric field is **perpendicular** to the equipotential line at that point, and it is directed toward lines of lower voltages.
- **b)** At any point, the electric field is **parallel** to the equipotential line at that point.
- **c)** At any point, the electric field is **perpendicular** to the equipotential line at that point, and it is directed toward lines of higher voltages.

**Correct statement:**  
The electric field is always **perpendicular** to equipotential lines and points from higher to lower potential (voltage).

**ANSWER:** a)

---

### **Part 3**

#### 4. Which statement best describes the distribution of the equipotential lines?

- **a)** The equipotential lines are equally spaced. The distance between each line is the same for all regions in the field.
- **b)** The equipotential lines are closer together in regions where the electric field is stronger.
- **c)** The equipotential lines are closer together in regions where the electric field is weaker.

**Correct statement:**  
Equipotential lines are **closer together where the electric field is stronger**, because the potential changes more rapidly in those regions.

**ANSWER:** b)

---

## **Final Answers**

**1.** $ 4.5 $ V  
**2.** $ 3.0 $ V  
**3.** a) At any point, the electric field is perpendicular to the equipotential line at that point, and it is directed toward lines of lower voltages.  
**4.** b) The equipotential lines are closer together in regions where the electric field is stronger.

VIPSolutions · Answer

Certainly! Here's a detailed, step-by-step explanation for each part, including the reasoning, formulas, and calculations:

---

## **Part 1: Voltage at different distances from a point charge**

### **Understanding the concept:**
- The electric potential $ V $ due to a point charge $ Q $ at a distance $ r $ is given by:
  
  $$
  V = \frac{kQ}{r}
  $$
  
  where $ k $ is Coulomb's constant ($ 8.99 \times 10^9 \, \mathrm{Nm^2/C^2} $).  
- Since $ V \propto \frac{1}{r} $, the potential decreases with increasing distance.

### **Given:**
- Voltage at 1 m from the charge, $ V_1 = 9\, V $.

### **Question 1: What is the voltage 2 m away?**

**Step 1: Write the proportionality:**

$$
V \propto \frac{1}{r}
$$

**Step 2: Set up the ratio:**

$$
\frac{V_2}{V_1} = \frac{r_1}{r_2}
$$

where:

- $ V_1 = 9\, V $ at $ r_1 = 1\, m $,
- $ r_2 = 2\, m $,
- $ V_2 = ? $.

**Step 3: Calculate $ V_2 $:**

$$
V_2 = V_1 \times \frac{r_1}{r_2} = 9\, V \times \frac{1\, m}{2\, m} = 9\, V \times 0.5 = 4.5\, V
$$

**Final answer:** **4.5 V**

---

### **Question 2: What is the voltage 3 m away?**

**Step 1:** Use the same proportionality:

$$
V_3 = V_1 \times \frac{r_1}{r_3} = 9\, V \times \frac{1\, m}{3\, m} = 9\, V \times \frac{1}{3} = 3\, V
$$

**Final answer:** **3.0 V**

---

## **Part 2: Relationship between electric field and equipotential lines**

### **Understanding:**

- **Electric field $ \vec{E} $** points in the direction of decreasing potential.
- **Equipotential lines** are lines where the potential $ V $ is constant.
- The electric field is **perpendicular** to these lines.
- The magnitude of the electric field relates to how rapidly the potential changes with distance.

### **Question 3: Which statement is true?**

Let's analyze each option:

**a)** *"At any point, the electric field is perpendicular to the equipotential line at that point, and it is directed toward lines of lower voltages."*

- **True.** The electric field always points from higher to lower potential and is perpendicular to equipotential lines.

**b)** *"At any point, the electric field is parallel to the equipotential line at that point."*

- **False.** Electric field is perpendicular, not parallel, to equipotential lines.

**c)** *"At any point, the electric field is perpendicular to the equipotential line at that point, and it is directed toward lines of higher voltages."*

- **False.** Electric field points **toward lower voltages**, not higher.

**Final answer:** **a)**

---

## **Part 3: Distribution of equipotential lines**

### **Understanding:**

- The potential difference between two equipotential lines is constant (e.g., 3 V).
- The spacing of these lines indicates the strength of the electric field:
  - **Closer lines** mean a **stronger electric field**.
  - **Farther lines** mean a **weaker electric field**.

### **Question 4: Which statement best describes the distribution?**

**a)** *"The equipotential lines are equally spaced. The distance between each line is the same for all regions."*

- **False.** In fields with non-uniform strength, spacing varies.

**b)** *"The equipotential lines are closer together in regions where the electric field is stronger."*

- **True.** The closer the lines, the stronger the electric field.

**c)** *"The equipotential lines are closer together in regions where the electric field is weak."*

- **False.** Closer spacing indicates stronger fields.

**Final answer:** **b)**

---

# **Summary of Final Answers:**

| Part | Question | Final Answer | Explanation |
|---------|-------------------------|-------------------|----------------------------------------------------------|
| 1 | Voltage at 2 m from charge | **4.5 V** | Voltage inversely proportional to distance $ V \propto 1/r $. |
| 2 | Voltage at 3 m from charge | **3.0 V** | Same proportionality, $ V \propto 1/r $. |
| 3 | Electric field and equipotential lines | **a)** | Electric field is perpendicular and points toward lower potential lines. |
| 4 | Distribution of equipotential lines | **b)** | Lines are closer in regions of stronger electric field. |

---

Feel free to ask if you'd like further clarification or more detailed derivations!

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