Section 21.4 Electric Field and Electric Forces
proton is placed in a uniform electric field of
. Calculate: (a) the magnitude of the electric force
felt by the proton; (b) the proton’s acceleration; © the proton’s
speed after s in the field, assuming it starts from rest.

Ask by Schmidt Pritchard. in the United States

Jan 14,2025

Question

Section 21.4 Electric Field and Electric Forces
proton is placed in a uniform electric field of
. Calculate: (a) the magnitude of the electric force
felt by the proton; (b) the proton’s acceleration; © the proton’s
speed after s in the field, assuming it starts from rest.

Ask by Schmidt Pritchard. in the United States

Jan 14,2025

UpStudy AI · Accepted Answer

Certainly! Let's solve each part of the problem step by step.

### **Given:**
- **Electric Field (E):** $$ 2.75 \times 10^{3} \, \text{N/C} $$
- **Charge of a Proton (q):** $$ 1.602 \times 10^{-19} \, \text{C} $$
- **Mass of a Proton (m):** $$ 1.673 \times 10^{-27} \, \text{kg} $$
- **Time (t):** $$ 1.00 \, \mu\text{s} = 1.00 \times 10^{-6} \, \text{s} $$

---

### **Part (a): Magnitude of the Electric Force Felt by the Proton**

**Formula:**
$$ F = qE $$

**Calculation:**
$$
F = (1.602 \times 10^{-19} \, \text{C}) \times (2.75 \times 10^{3} \, \text{N/C}) 
$$

$$
F = 1.602 \times 2.75 \times 10^{-19 + 3} \, \text{N}
$$

$$
F = 4.4055 \times 10^{-16} \, \text{N}
$$

**Rounded Answer:**
$$
F \approx 4.41 \times 10^{-16} \, \text{N}
$$

---

### **Part (b): Proton's Acceleration**

**Formula:**
$$ a = \frac{F}{m} $$

**Calculation:**
$$
a = \frac{4.4055 \times 10^{-16} \, \text{N}}{1.673 \times 10^{-27} \, \text{kg}}
$$

$$
a \approx 2.638 \times 10^{11} \, \text{m/s}^2
$$

**Rounded Answer:**
$$
a \approx 2.64 \times 10^{11} \, \text{m/s}^2
$$

---

### **Part (c): Proton's Speed After $$ 1.00 \, \mu\text{s} $$ in the Field**

**Assumption:**
- The proton starts from rest, so initial velocity ($$ v_0 $$) is 0.

**Formula:**
$$ v = v_0 + at $$
Since $$ v_0 = 0 $$, 
$$ v = at $$

**Calculation:**
$$
v = (2.638 \times 10^{11} \, \text{m/s}^2) \times (1.00 \times 10^{-6} \, \text{s})
$$

$$
v = 2.638 \times 10^{5} \, \text{m/s}
$$

**Rounded Answer:**
$$
v \approx 2.64 \times 10^{5} \, \text{m/s}
$$

---

### **Summary of Results:**

(a) **Electric Force:** $$ F \approx 4.41 \times 10^{-16} \, \text{N} $$

(b) **Acceleration:** $$ a \approx 2.64 \times 10^{11} \, \text{m/s}^2 $$

(c) **Speed After $$ 1.00 \, \mu\text{s} $$:** $$ v \approx 2.64 \times 10^{5} \, \text{m/s} $$
(a) The electric force on the proton is approximately $$ 4.41 \times 10^{-16} $$ N. (b) The proton's acceleration is approximately $$ 2.64 \times 10^{11} $$ m/s². (c) After $$ 1.00 \, \mu\text{s} $$, the proton's speed is approximately $$ 2.64 \times 10^{5} $$ m/s.

Section 21.4 Electric Field and Electric Forces
proton is placed in a uniform electric field of
. Calculate: (a) the magnitude of the electric force
felt by the proton; (b) the proton’s acceleration; © the proton’s
speed after s in the field, assuming it starts from rest.

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Section 21.4 Electric Field and Electric Forces proton is placed in a uniform electric field of . Calculate: (a) the magnitude of the electric force felt by the proton; (b) the proton’s acceleration; © the proton’s speed after s in the field, assuming it starts from rest.

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Latest Physics Questions

Section 21.4 Electric Field and Electric Forces
proton is placed in a uniform electric field of
. Calculate: (a) the magnitude of the electric force
felt by the proton; (b) the proton’s acceleration; © the proton’s
speed after s in the field, assuming it starts from rest.