Give step-by-step solution with explanation and final answer:on Was es J de LC Electric field Model
An Elect 1d is created by a +500nC charge particle. The electric field strength at an “” distance a

Question

Give step-by-step solution with explanation and final answer:on Was es J de LC Electric field Model
An Elect 1d is created by a +500nC charge particle. The electric field strength at an “” distance away
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A. Calculate the electric field strength at following points 1, 2 and 3? Draw the electric field vector
‘at each point corresponding to the magnitude and direction.
B. What is the electric force on a +8 nC charge if placed on those points? (direction and magnitude)
C. What is the electric force on a -5 nC charge if placed on those points? (direction and magnitude)
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VIPSolutions · Accepted Answer

Let's solve each part step by step:

---

## **Given Data**

- Charge creating the field: $ q = +500 \text{ nC} = +500 \times 10^{-9} \text{ C} $
- Electric field equation:  
  $$
  E = \frac{k |q|}{r^2}
  $$
- Coulomb's constant: $ k = 9. \times 10^9 \ \text{N} \cdot \text{m}^2/\text{C}^2 $
- Point 1: $ r_1 = .017 $ m  
- Point 2: $ r_2 = .01 $ m  
- Point 3: $ r_3 = .025 $ m

---

## **A. Calculate Electric Field Strength at Points 1, 2, and 3**

### **1. Point 1 ($ r_1 = .017 $ m):**

$$
E_1 = \frac{9. \times 10^9 \times 500 \times 10^{-9}}{(.017)^2}
$$

$$
E_1 = \frac{9. \times 10^9 \times 500 \times 10^{-9}}{.000289}
$$

$$
E_1 = \frac{450}{.000289}
$$

$$
E_1 \approx 1.56 \times 10^7 \ \text{N/C}
$$

---

### **2. Point 2 ($ r_2 = .01 $ m):**

$$
E_2 = \frac{9. \times 10^9 \times 500 \times 10^{-9}}{(.01)^2}
$$

$$
E_2 = \frac{450}{.0001}
$$

$$
E_2 = 4.5 \times 10^7 \ \text{N/C}
$$

---

### **3. Point 3 ($ r_3 = .025 $ m):**

$$
E_3 = \frac{9. \times 10^9 \times 500 \times 10^{-9}}{(.025)^2}
$$

$$
E_3 = \frac{450}{.000625}
$$

$$
E_3 = 7.2 \times 10^6 \ \text{N/C}
$$

---

**Direction:**  
The electric field points **radially outward** from the positive charge at all points.

---

## **B. Force on +8 nC Charge at Each Point**

$$
F = qE
$$
Where $ q = +8 \text{ nC} = 8 \times 10^{-9} \text{ C} $

### **Point 1:**  
$$
F_1 = 8 \times 10^{-9} \times 1.56 \times 10^7 = .125 \text{ N}
$$  
(Direction: Away from center)

### **Point 2:**  
$$
F_2 = 8 \times 10^{-9} \times 4.5 \times 10^7 = .36 \text{ N}
$$  
(Direction: Away from center)

### **Point 3:**  
$$
F_3 = 8 \times 10^{-9} \times 7.2 \times 10^6 = .0576 \text{ N}
$$  
(Direction: Away from center)

---

## **C. Force on -5 nC Charge at Each Point**

$$
q = -5 \text{ nC} = -5 \times 10^{-9} \text{ C}
$$

### **Point 1:**  
$$
F_1 = -5 \times 10^{-9} \times 1.56 \times 10^7 = -.078 \text{ N}
$$  
(Direction: Toward the center)

### **Point 2:**  
$$
F_2 = -5 \times 10^{-9} \times 4.5 \times 10^7 = -.225 \text{ N}
$$  
(Direction: Toward the center)

### **Point 3:**  
$$
F_3 = -5 \times 10^{-9} \times 7.2 \times 10^6 = -.036 \text{ N}
$$  
(Direction: Toward the center)

---

## **Final Answers Table**

| Point | $ r $ (m) | $ E $ (N/C)          | $ F_{8\,\text{nC}} $ (N) | $ F_{-5\,\text{nC}} $ (N) | Direction                  |
|-------|-------------|------------------------|----------------------------|-----------------------------|----------------------------|
| 1     | .017       | $ 1.56 \times 10^7 $ | .125                      | -.078                      | Outward for +, inward for -|
| 2     | .01        | $ 4.5 \times 10^7 $  | .36                       | -.225                      | Outward for +, inward for -|
| 3     | .025       | $ 7.2 \times 10^6 $  | .0576                     | -.036                      | Outward for +, inward for -|

---

Let me know if you need the field vectors drawn!

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