What is the percent uncertainty of the field? How does it compare to the electri
ID: 2078527 • Letter: W
Question
What is the percent uncertainty of the field? How does it compare to the electric field value estimated based on the positive and negative bar voltages? Which of the two is more accurate and why?
Workbook1 Q Search Sheet Insert Page Layout Formulas Data Review View +Share Store Bing Maps . Recommended A , ., , sparklines sicer mperl k Comment To t eader - PivotTable Recommended Table Pictures Shapes :@ Screenshot My Add-ins People Graph Comment Bx Box Footer 0 Average Std Dev Electric Field (N/C) 4.36 4.45 4.46 4.41 3.72 4.46 4.56 3.71 4.51 3.79 3.72 3.04 1.72 0.96 0.31 3.62 3.74 3.6 3.08 3.06 3.05 1.67 1.72 1.72 0.94 0.107 0.46 56.83 Average 3.34 Std Dev 10 Voltage 11 4.94V Distance x) 0.079m Electric Field (N/C) 62.5 N/C Seriess 4-5 2-3 1-2 0-1 1 2 345 6 01 2 13 Sheet +Explanation / Answer
the lines of equipotential have the same potential at a particular surface. hence voltages of the equipotential lines are consistent. E = (V+ - V-)/d, distance between bars is constant, hence electic field between positive and negative bar is constant. magnitude of electric field is as in above line and magnitude from positive to negative bar. The percentage error in electric field is almost negligible.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.