Understanding Potential Difference in Terms of Electric Field and Distance

Click For Summary
SUMMARY

The discussion centers on the calculation of electric potential difference (ΔV) in relation to electric field (E) and distance (Δx). The user correctly identifies that Δx equals 2 meters and ΔE equals 200 V/m. The confusion arises from the book's solution, which incorrectly interprets the vertical axis of the E vs. x graph as Volts instead of V/m, leading to an erroneous calculation of ΔV. The correct formula for ΔV is derived from the area under the E vs. x graph, confirming that ΔV should be calculated as -Δx * ΔE / 2, resulting in -200 V.

PREREQUISITES
  • Understanding of electric fields and potential difference
  • Familiarity with graph interpretation, specifically E vs. x graphs
  • Knowledge of basic calculus for area under curves
  • Concept of voltage as a measure of electric potential
NEXT STEPS
  • Study the relationship between electric field and potential difference in detail
  • Learn how to calculate areas under curves in physics, particularly in electric field contexts
  • Explore the implications of units in physics, focusing on V/m versus V
  • Review common misconceptions in electric field calculations and their resolutions
USEFUL FOR

Students of physics, educators teaching electromagnetism, and anyone seeking to clarify concepts related to electric fields and potential differences.

Miike012
Messages
1,009
Reaction score
0
I posted the question and the answer from the book inside the paint document.

My question is why is the answer not...

Δx = 3-1 = 2m
ΔE = 200-0 = 200V/m

ΔV = -(Area under E vs x graph) = -Δx*ΔE/2 = -200 V
 

Attachments

  • Question.jpg
    Question.jpg
    7.3 KB · Views: 400
Physics news on Phys.org
Miike012 said:
I posted the question and the answer from the book inside the paint document.

My question is why is the answer not...

Δx = 3-1 = 2m
ΔE = 200-0 = 200V/m

ΔV = -(Area under E vs x graph) = -Δx*ΔE/2 = -200 V
Your answer looks right to me.

What the book has for a solution is bizarre .
 
The vertical axis isn't Volts it's V/m (eg the rate of change of V with distance).
 

Similar threads

Replies
1
Views
2K
Replies
9
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
6
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
8
Views
2K
  • · Replies 6 ·
Replies
6
Views
5K
  • · Replies 22 ·
Replies
22
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
4
Views
6K