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- A charge of + 8 uC is evenly distributed over the volume of an insulating sphere of 6 cm radius. Gauss using the law of charge distribution center a) r = 4 cm and b) electric field at r = 8 cm distances find? c) Find the electric flux through a Gaussian surface given at r = 8 cm? NOTE: Englısh not my tongue language so sometimes i cant reading handwriting can u use computer ? thanksConsider a uniform electric field of E = (a, b, 0) and a disk of radius R What is the flux through the disk if it sits in the yz-plane? What is the flux through the disk if it sits in the xy-plane? What is the maximum flux through the disk as it rotated through all possible orientations?Determine the total charge distribution on r = b, and on r = a., explain Check if the electrical field for this configuration match the findings that for all points r inside the circle r<a is E= q/4piΕor2 and for all points r outside the shell r>a is E=0
- c. i. Compare the electric flux in a cubical surface of a side 10cm and spherical surface ofradius 10 cm, when a charge of 5µC is enclosed by them.ii. How does the electric field enclosing a given charge vary when the area enclosed bythe charged is doubled. Assuming the electric flux is constant.If the surface charge density on the surface of a conductor, σ = (786 – X) mC/m2 , (where X is your Roll No.). Apply Gauss’ Law to find (with complete logical an mathematical justification): the normal (perpendicular) component of the Electric Field exactly above and below the boundary of the surface if the conductor is placed in free space.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?
- A disk of radius 0.10 m is oriented with its normal vector n at 30 degrees to a unform electric field E vector of a magnitude of 2.0x10^3 N/C. What is the electric flux through the disk? (Provide the complete details for the Illustrated Diagram - inside the box, Given, Required, Equation, Solution, and Answer)A uniform electric field measured over a square surface with side length d = 15.5 cm makes an angle θ = 67.0° with a line normal to that surface, as shown in the figure below. There is a square horizontal surface with length and width d. The surface has a normal vertical axis at the center with a vector labeled vector E traveling from the center of the plane up and to the right. Vector E and the normal vertical axis form an angle labeled θ. If the net flux through the square is 5.40 N · m2/C, what is the magnitude Eof the electric field (in N/C)? N/CWhat is the area of the square sheet if an electric flux of is passing thru it, with an electric field of and creating an angle of 30° from the surface area of the square sheet? From the problem, which of the following is the correct variable for the required? Group of answer choices A E Φ φ
- A solid insulating plastic sphere of radius a carries atotal net positive charge 3Q uniformly distributed throughout its interior.The insulating sphere is coated with a metallic layer of inner radius a andouter radius 2a. The conducting metallic layer carries a net charge of -2Q. Apply Gauss’s law to find the magnitude of the electric field in the region r < a. Inthe figure, draw the Gaussian surface you are using, and indicate on that surface the direction of anyvectors which appear in the mathematical expression of Gauss’s law. Express your answer in terms ofa, Q, r, and ε0. (If you get an expression involving ρ, substitute it from above to re-express youranswer in terms of the stated variables.)Two infinite, nonconducting sheets of charge are parallel to each other . The sheet on the left has a uniform surface charge density σ, and the one on the right has a uniform charge density −σ. Calculate the electric field at points (a) to the left of, (b) in between, and (c) to the right of the two sheets. (d) What If? Find the electric fields in all three regions if both sheets have positive uniform surface charge densities of value σ.a loaded thin rod is bent into a semicircle of radius as shown in the figure. Charge density at a point on the rod, o the angle between the position vector of the point and the vertical c R λ = Acos(θ) : It is bound by the os expression. At the center of the semicircle (point O) what is the electric field?