d and e The field due to two parallel plates, the left with a negative surface charge density of -6.0 C/m2 while thesheet on the right has a positive surface charge density of +9.0C/m2 is given below. Suppose the plateswere spaced 0.0020 m apart. We found that, using Gauss's law that, independent of this distance,In I the net field points to the left and has magnitude |Enet| = 1.69 x 1011 N/C.• In region II, between the plates, the field has magnitude 8.47 x 1011 N/C and points to the left• In III it points to the right and has magnitude |Enet| = 1.69 x 1011 N/C.I1.1|||(a) Plot qualitatively correct E(x) and V(x) graphs, setting V=0 at the left (negatively charged) plate andletting x be the distance from the left plate, with positive x to the right. Use positive values for E if it points tothe right and negative if it points to the left. Your plot should include the range of x from -0.002 m to +0.004m. Check that the direction and size of the slope of your V graph is consistent with your E graph values.(b) In what direction would a positive test charge be pushed in each region? Discuss from the point of viewof the field and from the point of view of voltage.(c) What would happen in each region to a negative test charge moving to the right? (Would its speedincrease, decrease, remain the same?)(d) How would your graph change if you chose V=0 to be at the positive plate? Draw that graph.(e) What is the potential difference (AV) between the plates?

Given: density on left plate, σ1 = −6.0 C/m2density on right plate, σ2 = 9.0 C/m2distance between…

The field due to two parallel plates, the left with a negative surface charge density of -6.0 C/m2 while the sheet on the right has a positive surface charge density of +9.0C/m2 is given below. Suppose the plates were spaced 0.0020 m apart. We found that, using Gauss's law that, independent of this distance,

The field due to two parallel plates, the left with a negative surface charge density of -6.0 C/m2 while the sheet on the right has a positive surface charge density of +9.0C/m2 is given below. Suppose the plates were spaced 0.0020 m apart. We found that, using Gauss's law that, independent of this distance,

University Physics Volume 2

18th Edition

ISBN:9781938168161

Author:OpenStax

Publisher:OpenStax

Chapter6: Gauss's Law

Section: Chapter Questions

Problem 49P: Determine if approximate cylindrical symmetry holds for the following situations. State why or why...

See similar textbooks

Similar questions

The field due to two parallel plates, the left with a negative surface charge density of −6.0 C/m2 whilethe sheet on the right has a positive surface charge density of +9.0 C/m2is given below. Suppose the plates were spaced 0.0020 m apart. We found that, using Gauss’s law that, independent of this distance, • In I the net field points to the left and has magnitude |Enet| = 1.69 × 1011 N/C.• In region II, between the plates, the field has magnitude 8.47×1011 N/C and points to the left• In III it points to the right and has magnitude |Enet| = 1.69 × 1011 N/C.
The charge per unit length on the thin rod shownbelow is λ . What is the electric field at the point P? (Hint:Solve this problem by first considering the electric fieldd E→ at P due to a small segment dx of the rod, whichcontains charge dq = λdx . Then find the net field byintegrating d E→ over the length of the rod.)
The charge per unit length on the thin rod shown below is λ . What is the electric field at the point P? (Hint: Solve this problem by first considering the electric field dE at P due to a small segment dx of the rod, which contains charge dq = λdx . Then find the net field by integrating dE over the length of the rod.)
Show electrostatically that the electric field vector can be written as the gradient of a scalar.
As shown in Figure, a particle of charge +Q produces an electric field of magnitude Epart at point P, atdistance R from the particle. As in figure b, that same amount of charge is spread uniformly along a circulararc that has radius R and subtends an angle ?. The charge on the arc produces an electric field of magnitudeEarc at its center of curvature P. For what value of ? does Earc = 0.50 Epart ?
Figure 2 shows a nonconducting rod with a uniformly distributed charge Q. The rod forms a half-circle with radius R and produces an electric field of magnitude Earc at its center of curvature P. If the arc is collapsed to a point at distance R from P, by what factor is the magnitude of the electric field at P multiplied?
An infinite sheet of charge with surface charge density o = -87 C/m? lives in the x-y plane. A thin rod of charge with length L = 13.0 cm is placed along the +z-axis that the end closest to the sheet is located a distance d = 2.0cm away from the sheet. When held at this position, the linear charge density of the rod can be described by K =az, where a = +35 nC/m. 1, What is the force F on the rod from the sheet? 2, Can you draw a picture to represent it?
Find the electric field everywhere of an infinite uniform line charge with total charge Q. Sol. Using Gauss's Law: (contour integral E with rightwards harpoon with barb upwards on top) * (stack d A with rightwards harpoon with barb upwards on top) = ___________ /epsilon0 EA = ___________ / epsilon0 Since, its the line charge we use the area of a cylinder surrounding the line charge E*2pi ___________ L= ___________ / epsilon0 But all the charged get enclosed by the cylinder area, so qenc = Q Deriving, we get: E = (1/2(pi*epsilon0))*(___________ /rL) **see picture for the symbols used
Consider two thin disks, of negligible thickness, of radius R oriented perpendicular to the x axis such that the x axis runs through the center of each disk. The disk centered at x=0 has positive charge density η, and the disk centered at x=a has negative charge density −η, where the charge density is charge per unit area. What is the magnitude E of the electric field at the point on the x axis with x coordinate a/2? Express your answer in terms of η, R, a, and the permittivity of free space ϵ0.
1.Charge, Q= 2.20nC is uniformly distributed along the thin rodof length, L= 1.20m shown above. a.Determine the linear charge density of the rod. b.Determine an expression for charge dqon an infinitesimally small section of the rod of length dxin terms of the charge density and dx. c.The point P1is located a distance a= 0.300m from the end of the rod. Determine an expression for the distance, rbetween the point P1and an arbitrary chunk of the rod located an arbitrary distance xfrom the origin (between 0 and L). d.Determine an expression for the electric field dEat P1due to only the arbitrary chunk of the rod from part c. Use k, λ, x, L, and a. e.Determine the net electric field, Eat a point P1due to the entire rod.
According to Gauss’ law, when a long rod or wire is charged with a uniform linear charge density, the electric field due to the uniform line of charge is proportional to ______, where is the distance between the rod and the Gaussian surface. Select one: a. r2 b. 2r c. 1/r d. r
A solid ball of radius R has a uniform volume charge density and produces a certain electric field magnitude E1 at point P, a distance 3.08R from the ball's center. If a core of radius 0.385R is removed from the ball, what fraction of E1 will the field magnitude at P be?