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Fundamentals of Electromagnetics with Engineering Applications
- Given below, please solve for the total electric flux passing through the rectangular surface z=2, 0 < x < 2, 1 < y < 3, in the az direction. Please show your clean and complete solution. Thank youarrow_forwardThe cylindrical surface ρ = 8 cm contains the surface charge density, ρS = 5e−15|z| nC/m2. What is the total amount of charge present? How much electric flux leaves the surface ρ = 8 cm, 2 cm < z < 3 cm, 30° < φ < 60°?arrow_forwardHey I was wondering if you can help me with this problem plz Figure shows a plastic rod with a uniform charge −Q. It is bent in a 120° circular arc of radius r and symmetrically placed across an x axis with the origin at the center of curvature P of the rod. In terms of Q and r, what is the electric field E ⃗ due to the rod at point P?arrow_forward
- (b) An infinite nonconducting sheet has a surface charge density sigma= 0.30x10^-6 C/m^2 on one side. How far apart are equipotential surfaces whose potentials differ by 100 V?arrow_forwardIf a system of charge is formed by a vertical stack of infinite parallel planes each with a uniformsurface charge density, what is the relation of the electric field above the top plane to the electric field below thebottom plane? (a) This cannot be determined without knowing the charges of the individual planes.(b) equal in magnitude, opposite in direction(c) equal in magnitude and in direction(d) The field above the top plane is larger than the field below the bottom plane; the fields point in oppositedirections.arrow_forwardA rectangular flat surface with sides 0.200 m and 0.500 m is under the influence of a uniform electric field E=850 that is directed at 20.0" from the plane of the rectangular sheet. Find the electric flux through the rectangular fat sheet.arrow_forward
- Determine the work done, in µJ, in moving a 8-nC charge from A(3, 4, 3) m to B(-3, 3, -1) m against the electric field due to an infinite sheet charge of density 6 nC/m^2 at y = -1 m.arrow_forwardUse Gauss’s law to derive the expression for the electric field between two uniformly charged large parallel sheets with surface charge densities a and -a respectively.arrow_forwardThree infinite uniform sheets of charge are located in free space as follows: 3 nC/m2 at z = − 4, 6 nC/m2 at z = 1, and − 8 nC/m2 at z = 4. Find E at the points: (a) PA(2, 5, − 5); (b) PB(4, 2, −3); (c) PC(−1, −5, 2); (d) PD(−2, 4, 5).arrow_forward
- Please write to text format The charge per unit length on the thin rod shown below is ?. 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 = ? dx. Then find the net field by integrating dE over the length of the rod. Use the following as necessary: L, a, ?, and ?0. Enter the magnitude. Assume that ? is positive.) Earrow_forwarda long straight cylindrical wire of radius r meter, in a medium of permittivity e is parallel to a horizontal plane conducting sheet. The axis of the wire is it expr metres above the sheet (a) Derive an expression of the capacitance per unit length between the wire and the sheet (b) If r = 0.3 x 10-2 m, h.= 0.12 m find the capacitance per metre length (c) If the potential difference betweenthe wire and sheet is 5 kV, find the magnitude and direction of electric stress in the medium at theupper surface of the sheet at a distance 20 cm from the axis of the wire. Take e = 1/36π x 10-9 F/m [(a) C = 2πe/ln 2h - r/r F/m (b) 0.0127 x 10-9 F/rn (c) 6.85 kV/m acting vertically downward]arrow_forwardGiven a charge distribution with density ρv = 5r C/m3 in spherical coordinates, use Gauss Law to find D.arrow_forward
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