M1.2-answer number 5 only -complete solution PDF Activity 1.2 - Electric Flux and Ga x+O File | C:/Users/Danica%20Martinez/Desktop/Activity%201.2%20-%20Electric%20Flux%20and%20Gauss's%20Law(1).pdf+1...a Agoda G GmailYouTube G Maps O Translate E MODULE 4.2 Electr.INSTRUCTIONS:Show your complete solution to each problem on a short bond paper.Answers must be expressed in engineering notation (when the exponent of the base 10 multiplieris not a multiple of 3, press ENG or SHIFT+ENG, whichever the case.)Example: 0.06N or 6.0x10-2 N must be expressed to 60x10-3 N or 60mN1. A hemispherical surface with radiusr = 12 cm in a region of uniform electric field E = 10 V/m has itsaxis aligned parallel to the direction of the field. Calculate the flux through the surface.2. In the figure shown, the magnitude of the uniform electricfield is E = 70 kN/C. Calculate the electric flux through:320 cmA. The vertical rectangular surfaceB. The slanted surfaceC. The entire surface of the box45.0 cm3. Consider a thin, spherical shell of radius 12.0 cm with a surface charge density of 0.150 mC/m2distributed uniformly on its surface. Find the electric field:A. 7.0 cmB. 17.0 cm from the center of the charge distribution.4. Four small spheres shown in the figure below have chargesof g1 = -3.00 nC, q2 = 2.40 nC, g3 = 5.30 nC, and g4 = -1.80nC. Find the net electric flux through each of the followingclosed surfaces:S2S,A. S,B. S2C. SaD. S492S45. The electric field at a distance of 0.35 m from a very long line charge is 1250 N/C. Calculate thetotal charge in a 6.00-cm section of such line.11:07 amP Type here to search32°C HazeA ENG19/03/2022

Given:Electric field, E = 1250 N/Cdistance, r = 0.35 mlength, l = 6 cm = 6×10-2 mcharge is 'q'

Answered: 5. The electric field at a distance of…

5. The electric field at a distance of 0.35 m from a very long line charge is 1250 N/C. Calculate the total charge in a 6.00-cm section of such line.

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

See similar textbooks

Related questions

Q: The hole density in an n-type Si slab (doped at Na = 10¹8 cm-3) decreases linearly from 10¹4 cm to…

Q: What is the drift velocity in n electric field of 1x10^5 Volts/cm?

Q: A 1200 resistor and a 500N resistor are connected in series. A 12V battery is hooked up to the…

A: Given : Ra = 120 Ω Rb = 500 Ω V = 12 V

Q: You are serving as an expert witness for the city council of a community. The council is exploring…

A: The community requires 2.20 MW of power for the whole day (24 hours).The best photovoltaic cells on…

Q: a) What is Hall Effect? Explain briefly.

A: Given : To explain hall effect.

Q: hello what sort of calculator did you use? i am using ti 84 and ti 89 still not getting the answer…

Q: For a science experiment you need to electroplate zinc onto both sides of a very thin, 3.00 cm x3.00…

Q: a) Calculate the electron scattering rate and the mean free path of copper at 295 K.

Q: A sample of n-type germanium (Ge) contains 1023 ionised donors per cubic metre. Estimate the ratio…

Q: What observable characteristic is common among all the solutions? What substance provides this…

A: Given : What are the common observable characteristic of solution.

Q: 3. Consider a two-cell series-connected battery pack where cell states of charge are z1 = 0.99 and…

A: Given, For cell 1, z1=0.99Q1=31Ah For cell 2, z2=0.97Q2=32Ah zmin=0.1 OCV=3.3V So present charge…

Q: What is the number density of free electron carries in the metallic element nickel if the electrons…

A: atomic mass of nickel = 58.6934 g/mol density of nickel = 8.902 gcm3number of conduction electrons…

Q: My professor gave me 5 multiple choice answers for this problem. Does any of these fit the answer…

A: Page 1

Q: 1.15 When a car has a dead battery, it can often be started by connecting the battery from another…

A: Given: v=12vI=40At=1.5min=90s

Q: Problems Q1: Calculate the drift current density in silicon sample. If T=300 K, Na-102¹/m³,…

Concept explainers

Electric Charges and Fields

One of the fundamental properties is the electromagnetic property. Charge cannot be destroyed by any process and this contributes formally to the law of charge conservation.

Question

M1.2
-answer number 5 only
-complete solution

PDF Activity 1.2 - Electric Flux and Ga x
+
O File | C:/Users/Danica%20Martinez/Desktop/Activity%201.2%20-%20Electric%20Flux%20and%20Gauss's%20Law(1).pdf
+1
...
a Agoda G Gmail
YouTube G Maps O Translate E MODULE 4.2 Electr.
INSTRUCTIONS:
Show your complete solution to each problem on a short bond paper.
Answers must be expressed in engineering notation (when the exponent of the base 10 multiplier
is not a multiple of 3, press ENG or SHIFT+ENG, whichever the case.)
Example: 0.06N or 6.0x10-2 N must be expressed to 60x10-3 N or 60mN
1. A hemispherical surface with radiusr = 12 cm in a region of uniform electric field E = 10 V/m has its
axis aligned parallel to the direction of the field. Calculate the flux through the surface.
2. In the figure shown, the magnitude of the uniform electric
field is E = 70 kN/C. Calculate the electric flux through:
320 cm
A. The vertical rectangular surface
B. The slanted surface
C. The entire surface of the box
45.0 cm
3. Consider a thin, spherical shell of radius 12.0 cm with a surface charge density of 0.150 mC/m2
distributed uniformly on its surface. Find the electric field:
A. 7.0 cm
B. 17.0 cm from the center of the charge distribution.
4. Four small spheres shown in the figure below have charges
of g1 = -3.00 nC, q2 = 2.40 nC, g3 = 5.30 nC, and g4 = -1.80
nC. Find the net electric flux through each of the following
closed surfaces:
S2
S,
A. S,
B. S2
C. Sa
D. S4
92
S4
5. The electric field at a distance of 0.35 m from a very long line charge is 1250 N/C. Calculate the
total charge in a 6.00-cm section of such line.
11:07 am
P Type here to search
32°C Haze
A ENG
19/03/2022

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.

Similar questions

For a science experiment you need to electroplate a 110-nm-thick zinc coating onto both sides of a very thin, 1.0 cm x 1.0 cm copper sheet. You know that the charge carriers in the ionic solution are divalent (charge 2e) zinc ions. The density of zinc is 7140 kg/m³. Part A If the electroplating apparatus operates at 1.7 mA, how long will it take the zinc to reach the desired thickness? Express your answer in seconds. G| ΑΣΦ t = Submit Previous Answers Request Answer ? X Incorrect; Try Again; 2 attempts remaining S
按ESC | 即可退出全屏模式 Gas insulation is important for transmission lines, outdoor insulators and Gas-insulated- Substations (GIS). Understanding the breakdown mechanisms and behaviour of insulation materials is key to the optimum design of insulation systems. The speed distribution function of electrons in a gas containing electrons, ions and atoms is controlled by the electric field strength, E, and can be expressed as fo (E, v). It satisfies the relationship of (E, v) dv=1. i) ii) Convert the above electron speed distribution function into an electron energy distribution function that satisfies the relationship f(E, ɛ) dɛ =1 where & is the electron energy and ƒ(E, ɛ) the electron energy distribution function. The electron number density is ne and the atom number density is na in the gas. The ionisation cross section of the atoms by electron impact is a function of electron energy, ɛ, and can be written as σ¡ (ɛ). Similarly, the attachment collision cross section between an electron and an…
Q2/ If the resistivity of pure silicon is 2300 (0. m) and assuming that the valance electrons of silicon is 4. Let T = 300 K. (a) Calculate the electron density per cubic meter (b) Calculate the electron mobility when the electron density is doubled. Note: Avogadro's number is 8.61 x 10 mol Atomic mass of silicon is 28.08 g/mol, mass density is 2.33 g/cm
Q1. Find the following: a. ZT b. VR C. V d. IT VR VE 10 Q X = 20 2 15 V/0° Vo TX = 70 Xo = 30
Solve. with me 30 minutes
For problem 24a (part a), what would the velocity of the electron (in units of 10^7 m/s) be if the electron were accelerated by a voltage difference of 338 V? (5 sig figs)
Question-5. A semiconductor has an electrical conductivity of 20 (N-m)', whereas the electron and hole mobilities are 0.04 and 0.03 m2/V-s, respectively. The density of the semiconductor is 4.62 g/cm³. The electrical charge of an electron (e) is 1.6 ×10-19 C. The atomic weight of the semiconductor is 59.72 g/mol. Avogadro constant (NA) is 6.023 × 1023 atoms/mol. (a) Compute the intrinsic carrier concentration for the semiconductor at room temperature (25 °C). (b) Compute the number of free electrons per atom for the intrinsic semiconductor at room temperature.
1. A certain geophysical technique determines the electrical resistivity of the ground by measuring the voltage, V, and current, I, between electrodes separated by a distance, a. These measurements are related to resistivity by the equation, resistivity = 2za (VI). 1.00 +0.01 meters, what If V = 0.512 +0.01 volts, I–0.00684 ± 0.0001 Amps, and a = is the uncertainty in the resistivity. (Note: The units of resistivity are ohm-meters. The numbers in this problem are given in such a way that no unit conversions are necessary.)
Equation1: Q(t)=CVcap(t) Equation 2: Qcharging(t)=CV(1−e^(−t/RC)) 1. Combine equations 1 and 2 to create an equation capable of finding the time-dependent voltage across a charging capacitor. Equation 3: I(t)≡(dQ(t))/(dt) 2. Combine equations 2 and 3 to create an equation capable of finding the time-dependent current across a charging capacitor.
Q. The ends of the cylinder are made of conducting discs and the space between these discs are filled with an inhomogeneous ohmic medium whose conductivity o = L/(z+1). Where L is the separation distance of the discs. A d-c voltage Vo is applied across the discs as shown on the right. Determine V. a) The total resistance between the discs b) The surface charge densities on the discs c) The volume charge density and the total amount of charge between the discs
The current–voltage characteristic curve for a semiconductor diode as a function of temperature T is given by I = I0(eeΔV/kBT - 1)Here the first symbol e represents Euler’s number, the base of natural logarithms. The second e is the magnitude of the electron charge, the kB stands for Boltzmann’s constant, and T is the absolute temperature. (a) Set up a spreadsheet to calculate I and R = ΔV/I for ΔV = 0.400 V to 0.600 Vin increments of 0.005 V. Assume I0 = 1.00 nA. (b) Plot R versus ΔV for T = 280 K, 300 K, and 320 K.
(a) How many silicon atoms are there in each unit cell? (b) How many silicon atoms are there in one cubic centimeter? (c) Knowing that the length of a side of the unit cell (the silicon lattice constant) is 5.43 Å, Si atomic weight is 28.1, and the Avogdaro's number is 6.02 × 10²3 atoms/mole, find the silicon density in g/cm³.