This question explores the difference between the integral • î dA over a closed Gaussian surface and the integral ·d i around a closed path. The electric field due to stationary charges (not shown) is measured at locations on a Gaussian box with dimensions L = 5 mm and h = w = 1.5 mm as shown below. The electric field on the surface of the right half of the box is uniform with magnitude 4 x 10° V/m. On the left half of the box E ~ 0. L/2 L/2 h Part 1 What is the total electric flux on the surface of the Gaussian box? i V-m Hint Assistance Used Save for Later Attempts: 0 of 10 used Submit Answer Part 2 What is the charge inside the box? 2 Tinside = C i Hint Save for Later Attempts: 0 of 10 used Submit Answer

Modern Physics
3rd Edition
ISBN:9781111794378
Author:Raymond A. Serway, Clement J. Moses, Curt A. Moyer
Publisher:Raymond A. Serway, Clement J. Moses, Curt A. Moyer
Chapter2: Relativity Ii
Section: Chapter Questions
Problem 4P: A charged particle moves along a straight line in a uniform electric field E with a speed v. If the...
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This question explores the difference between the integral
îî dA over a closed Gaussian surface and the integral
O É ·d i around a closed path.
The electric field due to stationary charges (not shown) is measured at locations on a Gaussian box with dimensions L = 5 mm and
h = w = 1.5 mm as shown below. The electric field on the surface of the right half of the box is uniform with magnitude 4 x 10° V/m.
On the left half of the box E 0.
L/2
L/2
Part 1
What is the total electric flux on the surface of the Gaussian box?
i
V-m
Hint
Assistance Used
Save for Later
Attempts: 0 of 10 used
Submit Answer
Part 2
What is the charge inside the box?
Σ
Tinside =
i
Hint
Save for Later
Attempts: 0 of 10 used
Submit Answer
Transcribed Image Text:This question explores the difference between the integral îî dA over a closed Gaussian surface and the integral O É ·d i around a closed path. The electric field due to stationary charges (not shown) is measured at locations on a Gaussian box with dimensions L = 5 mm and h = w = 1.5 mm as shown below. The electric field on the surface of the right half of the box is uniform with magnitude 4 x 10° V/m. On the left half of the box E 0. L/2 L/2 Part 1 What is the total electric flux on the surface of the Gaussian box? i V-m Hint Assistance Used Save for Later Attempts: 0 of 10 used Submit Answer Part 2 What is the charge inside the box? Σ Tinside = i Hint Save for Later Attempts: 0 of 10 used Submit Answer
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