Find the electric flux ( in units of N.m2/C) through the surface of the rectangle ( 30 cm × 40 cm) in the figure given that E = 720.4 N/C. Normal E 0=30°
Q: A flat square sheet of thin aluminum foil, 25.0 cm on a side, carries 275 nC of charge, which is…
A:
Q: Consider the uniform electric field E = (6.0j + 2.0 k) X 103 N/C. What is its electric flux (in N.…
A:
Q: Calculate the electric flux for each of the closed surfaces a, b, c, and d. d. 92 93 97 94 95 98 96…
A: From Gauss's law, the electric flux through a closed surface can be given by, φ=qencε0, where qenc…
Q: The total electric flux from a cubical box of side 28.0 cm is 1.85 x 103 Nm2/C. What charge is…
A: GivenSide of the box =28.0 cmTotal electric flux through the box (ϕ) = 1.85 × 103 Nm2/C
Q: A nonuniform electric field is given by the expression D = 2у a + z a — Зха y C/m?. Determine the…
A: We have Electric field given by D=2yi^+zj^-3xk^. We need to find Electric flux through a rectangular…
Q: A closed surface with dimensions a=b%3D0.40 m and c=0.60 m is located as in the figure below. The…
A:
Q: An electric field Ē = axî + By txzk exists, in space where.a, B.Y are constant. Electric flux…
A:
Q: A flat surface of area 1.25 m2 is rotated through a uniform horizontal electric field of 4.20 N/C.…
A:
Q: Find the electric flux through the plane surface shown in the figure below if = 61.8°, E = 372 N/C,…
A: We have a plane square surface of side d=4.7 cm=0.047 m through which an electric field of magnitude…
Q: Find the total electric flux over the horizontal cylinder of base area A. The electric field is…
A:
Q: The uniform surface charge density ps nC/m² is presented on the plane z = 15cm. Find the electric…
A: We know that the Electric field due to the surface charge 'es' will be equal to…
Q: The electric flux through an arbitrary closed surafce is ϕ=52Nm2/C. What is the charge enclosed by…
A:
Q: Consider a closed triangular box resting within a horizontal electric field of magnitude E = 7.54 x…
A: Given: Now,
Q: A cylinder of diameter 1.72 m is in a region where the electric field is as shown in the figure…
A: Flux is the cross product of electric field and cross section area.
Q: Find the electric flux ( in units of N.m²/C) through the surface in the figure ( dashed line) given…
A: Given values:- Charge,(Q)=35.6 pC=35.6 x 10-12 C -----(1) The permittivity of free space,(ε0)=8.8542…
Q: The electric flux through the shaded surface is А. 0 В. 200 N m/C 2 m С. 400 N m²/C D. Flux isn't…
A: Calculate the cross-sectional area of the shaded surface.
Q: Q.1 : How we can calculate flux in a closed sphere and charge is enclosed in it? Q.2 : Find the…
A: Electric flux=∫E→.ds→ where E→=electric field ds=small area element According to Gauss law…
Q: Calculate the electric flux D for each of the closed surfaces a, b, c, and d. b. 92. a 93 97 94 95…
A: According to Gauss law , electric flux from a closed surface is given by charged enclosed divided by…
Q: A 4.0 cm^2 in the x-y plane sits in aunifo electric field E=(2.0i+3.0j+5.0k)N/C. Find the electric…
A: Given: Area of the square in x-y plane, A = 4.0 cm2. Electric field, E = (2.0i+3.0j+5.0k) N/C. To…
Q: A circular surface with a radius of 0.063 m is exposed to a uniform electric field of magnitude 1.76…
A: Given that :E=1.76 x 104 N/Cr=0.063 mϕ=58 Nm2/Cwhere,E is the electric fieldr is the radius of the…
Q: 1. Find the electric flux through the surface with sides of 15 cm X 15 cm shown in the figure below.…
A: The answers are below
Q: Find the electric flux ( in units of N.m2/C) through the surface of the rectangle ( 30 cm \( \times…
A: Given: The electric field is E=720.4 N/C. The angle is θ=30°. The area is A=30 cm×40 cm. The…
Q: An electric flux of 144 N.m2/C passes through a flat horizontal surface that has an area of 0.72 m2.…
A: Electric flux through a surface (∅) = 144 N.m2 mCArea of surface (A) =0.72 m2Angle made by electric…
Q: A 3.7 cm × 4.0 cm rectangle lies in the xy-plane. What is the electric flux through the rectangle…
A: Given:
Q: A uniform electric field with a magnitude of 1.25x10^5 N/C passes through a rectangle with sides of…
A: Given: magnitude of electric field, E = 1.25 ×105 N/C Dimension of rectangle: 2.5 m × 2 m Electric…
Q: he electric flux (Φ) that passes through this surface
A: we know electric flux ϕ = E·A =EAcosθ where θ= angle between electric field and area vector A = area…
Q: A closed surface with dimensions a=b=0.40 m and c=0.60 m is located as in the figure below. The left…
A:
Q: A flat surface with an area of 1.25 m² is rotated through a uniform horizontal electric field of 5…
A:
Q: Find the electric flux ( in units of N.m2/C) through the surface of the rectangle ( 30 cm x 40 cm)…
A: We know that the Electric flux(∅) through surface of rectangle having area(A) and Electric field(E)…
Q: Consider a plane surface in a uniform electric field as shown in the figure, where d =20 cm and the…
A: The electric field is an area within which when a charged particle enters, a force gets exerted on…
Q: A 3.65x1012 Cis placed inside a cube. What is the amount of electric flux passing through each side?…
A:
Q: What is the flux due to electric field E = 3 × 103,NC−1 through side of the square 10cm , when it is…
A: Given : E = 3 × 103 NC–1 Side of square (S) = 10 cm = 0.1 m. Area of square (A) = (side)2 = (0.1)2 =…
Q: Consider a closed triangular box resting within a horizontal electric field of magnitude E = 6.86 ×…
A:
Q: A closed surface with dimensions a=b%=D0.40 m and c=0.60 m is located as in the figure below. The…
A: The electric flux is given by the product of surface area and the component of electric field…
Q: Find the total electric flux over the horizontal cylinder of base area A. The electric field is…
A:
Q: Draw the problem and answer, thanks. What is the area of a rectangle whose electric flux is 28…
A: Electric flux is defined as ∅=∫E→·dS→ Here E→ is the electric field and ∅ is flux and dS→ is…
Q: Find the electric flux ( in units of N.m2/C) through the surface of the rectangle (30 cm x 40 cm) in…
A: Write the given values of the problem- Electric field(E) E=224.8 N/C Suraface of the rectangle=30…
Q: An electric field of intensity 3.50 kN/C is applied along the x-axis. Calculate the electric flux…
A: Given data: Intensity of electric field is, E=3.50 kN/C =3.50×103 N/C. Width of rectangular plane…
Q: What is the total flux passing through a 10 cm × 6 cm surface in a region where the electric flux…
A: Electric flux is defined as the product of the electric field and the specified area. Electric flux…
Q: The flux ϕE=21N⋅m2/C through the open surface A⃗ =7i^+11j^+6k^ is given. Find the value of a if the…
A:
Q: A uniform electric field of strength E = 334 N/C passes through a flat surface at an angle of 0.880…
A: To find the area of the surface in given conditions. We are given with E=334 N/C angle θ=0.880…
Q: Find the electric flux ( in units of N.m2/C) through the surface of the rectangle ( 30 cm × 40 cm)…
A:
Q: A uniform electric field with a magnitude of 1.25x10° N/C passes through a rectangle with sides of…
A:
Q: A flat sheet is in the shape of a rectangle with sides of lengths 0.400 m and 0.600 m. The sheet is…
A: Given in question - length (L) = 0.600 m Width(B) = 0.400m…
Q: Find the electric flux ( in units of N.m?/C) through the surface of the rectangle ( 30 cm × 40 cm)…
A:
Step by step
Solved in 2 steps with 2 images
- The electric field that comes out perpendicularly from the surface of the circular crown of internal radii a=10 www um www wwwwwwwww wwwwwwwwww cm, external b=20 cm, depends on the radial distance "r", according to: www E=E.(r/rm)² Where Eo=500 N/C is a constant and "rm" is the mean radius of the crown. Find the electric flux (in Nom²/c²) passing through the surface of the crown. a) 50,4 b) 51,4 a 0 E c) 52,4 b d) 53,4 e) 54,4Find the electric flux (in units of N.m2/C) through the surface of the rectangle ( 30 cm x 40 cm) in the figure given that E = 399.7 N/C. Normal E 0-30What is the total flux passing through a 10 cm × 6 cm surface in a regionwhere the electric flux density is 2700 μC/m2?
- D3.3. Given the electric flux density, D = 0.3r²a, nC/m² in free space: (a) find E at point P(r = 2, 0 = 25°, ø = 90°); (b) find the total charge within the sphere r = 3; (c) find the total electric flux leaving the sphere r = 4. Ans. 135.5a, V/m; 305 nC; 965 nCThe electric flux through a spherical surface is 4.0104 N m2/C. What is the net charge enclosed by the surface?Find the electric flux through a rectangular area 3 cm x 2 cm between parallel plates where there is a constant electric field of 30 N/C for the following orientations of the area: (a) parallel to the plates, (b) perpendicular to the plates, and (c) the normal to the area making a 300 angle with the direction of the electric field. Note that this angle can also be given as 180o + 30o.
- A uniform electric field of magnitude 1.1104 N/C is perpendicular to a square sheet with sides 2.0 m long. What is the electric flux through the sheet?The electric flux through a cubical box 8.0 cm on aside is 1.2103 N m2/C. What is the total charge enclosed by the box?A square surface of area 2 cm2 is in a space of uniform electric field of magnitude 103 N/C . The amount of flux through it depends on how the square is oriented relative to the direction of the electric field. Find the electric flux through the square, when the normal to it makes the following angles with electric field: (a) 30(, (b) 90(, and (c) 0(. Note that these angles can also be given as 1800 + .
- The volume charge density of a spherically charged cloud changes with the distance to the cloud center as p(r) = po r a " where the parameters are po 1.85 nC/m³ and a = 22 m. Find the magnitude of electric field at distance d = 25 m from the cloud center. The electric field, E(r = 25 m) = 1484.65 Find the electric field (in N/C) as a function of r (in m). Do not submit the units. The electric field, E(r) = 1484.65 N/C. = Units N/CA uniform electric field of magnitude 5.7 x 104 N/C is at an angle of 10° to a square sheet with sides 5.5 m long. What is the electric flux through the sheet? Hint Electric flux is E N.m²/C.Two uniformly charged rods, bent into circular arcs, have equal and opposite linear chargedensities ±λ. Their ends are placed together so that they form a semicircle of radius R asshown in the Fig. Q1. What is the electric field strength at the center?