12) How much work must be done to move a proton from point A, at a potential of –75V, to point B, which is at a potential of +25 V, along the curved path shown in the figure? Assume the system is isolated from any outside forces. (charge on a proton = 1.60 × 10¬19 C) -75V +25V A) 1.6 J B) 1.60 × 10-17 J C)-1.60 x 10-15 J D) –1.6 J E) Need to know the length of the curved path
12) How much work must be done to move a proton from point A, at a potential of –75V, to point B, which is at a potential of +25 V, along the curved path shown in the figure? Assume the system is isolated from any outside forces. (charge on a proton = 1.60 × 10¬19 C) -75V +25V A) 1.6 J B) 1.60 × 10-17 J C)-1.60 x 10-15 J D) –1.6 J E) Need to know the length of the curved path
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![### Physics Problem: Work Done on a Proton
#### Problem Statement:
**Q12)** How much work must be done to move a proton from point A, at a potential of -75V, to point B, which is at a potential of +25V, along the curved path shown in the figure? Assume the system is isolated from any outside forces. (Charge on a proton = 1.60 x 10⁻¹⁹ C)
#### Options:
- **A)** 1.6 J
- **B)** 1.60 x 10⁻¹⁷ J
- **C)** -1.60 x 10⁻¹⁵ J
- **D)** -1.6 J
- **E)** Need to know the length of the curved path
#### Diagram Description:
- The diagram shows a path from point A, which is at a potential of -75V, to point B, at a potential of +25V.
- The path is curved and indicated by an arrow moving from left (point A) to right (point B).
---
To solve this problem, we need to calculate the work done on the proton as it moves from point A to point B.
**Solution Approach:**
The work done \(W\) in moving a charge \(q\) through a potential difference \(\Delta V\) is given by:
\[ W = q \Delta V \]
Where:
- \( q \) is the charge of the proton (1.60 x 10⁻¹⁹ C)
- \(\Delta V\) is the potential difference between the two points.
Therefore, the potential difference \(\Delta V\) is:
\[ \Delta V = V_B - V_A = 25V - (-75V) = 25V + 75V = 100V \]
Now, calculate the work done \(W\):
\[ W = (1.60 \times 10^{-19} \text{ C}) \times (100 \text{ V}) \]
\[ W = 1.60 \times 10^{-17} \text{ J} \]
From the given options, the correct answer is **B) 1.60 x 10⁻¹⁷ J**.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5aab90fe-f662-4329-be2f-bc153bcfca52%2F67e66c42-1461-4ccd-b413-7363c1b5a005%2Fbl5zlla_processed.png&w=3840&q=75)
Transcribed Image Text:### Physics Problem: Work Done on a Proton
#### Problem Statement:
**Q12)** How much work must be done to move a proton from point A, at a potential of -75V, to point B, which is at a potential of +25V, along the curved path shown in the figure? Assume the system is isolated from any outside forces. (Charge on a proton = 1.60 x 10⁻¹⁹ C)
#### Options:
- **A)** 1.6 J
- **B)** 1.60 x 10⁻¹⁷ J
- **C)** -1.60 x 10⁻¹⁵ J
- **D)** -1.6 J
- **E)** Need to know the length of the curved path
#### Diagram Description:
- The diagram shows a path from point A, which is at a potential of -75V, to point B, at a potential of +25V.
- The path is curved and indicated by an arrow moving from left (point A) to right (point B).
---
To solve this problem, we need to calculate the work done on the proton as it moves from point A to point B.
**Solution Approach:**
The work done \(W\) in moving a charge \(q\) through a potential difference \(\Delta V\) is given by:
\[ W = q \Delta V \]
Where:
- \( q \) is the charge of the proton (1.60 x 10⁻¹⁹ C)
- \(\Delta V\) is the potential difference between the two points.
Therefore, the potential difference \(\Delta V\) is:
\[ \Delta V = V_B - V_A = 25V - (-75V) = 25V + 75V = 100V \]
Now, calculate the work done \(W\):
\[ W = (1.60 \times 10^{-19} \text{ C}) \times (100 \text{ V}) \]
\[ W = 1.60 \times 10^{-17} \text{ J} \]
From the given options, the correct answer is **B) 1.60 x 10⁻¹⁷ J**.
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