The electric flux through the surface shown in FIGURE EX24.10 is 25 N m²/C . What is the electric field strength?

A flat sheet of paper of area 0.250 m2 is oriented so that the normal to the sheet is at an angle of 60° to a uniform electric field of magnitude 14 N/C. (a) Find the magnitude of the electric flux through the sheet. (b) Does the answer to part (a) depend on the shape of the sheet? Why or why not?

A 2.0 cm×3.0 cm rectangle lies in the 𝓍𝒵-plane with unit vector nˆ pointing in the +y-direction. What is the electric flux through the rectangle if the electric field is(b) E (→ above E) = (4000 î−2000 kˆ) N/C?

A 12 cm×12 cm rectangle lies in the first quadrant of the xy-plane with one corner at the origin. Unit vector nˆ points in the +𝒵 -direction. What is the electric flux through the rectangle if the electric field is E (→ above E) = (2000 m¯¹)x kˆ N/C?Hint: Divide the rectangle into narrow strips of width .

A 10 nC charge is at the center of a 2.0 m x 2.0 m x 2.0 m cube. What is the electric flux through the top surface of the cube?

Find the electric fluxes ΦA to ΦE through surfaces A to E in FIGURE P24.29.

Charges q₁ = -4Q and q2 = +2Q are located at 𝓍 = -a and 𝓍 = + a, respectively. What is the net electric flux through a sphere of radius 2a centered (a) at the origin and (b) at 𝓍 = 2a?

FIGURE P31.38 shows the electric field inside a cylinder of radius R=3.0 mm. The field strength is increasing with time as E=1.0×10^8t^2 V/m, where t is in s. The electric field outside the cylinder is always zero, and the field inside the cylinder was zero for t<0.a. Find an expression for the electric flux Φₑ through the entire cylinder as a function of time.

All examples of Gauss’s law have used highly symmetric surfaces where the flux integral is either zero or EA. Yet we’ve claimed that the net Φₑ = Qᵢₙ / ϵ₀ is independent of the surface. This is worth checking. FIGURE CP24.57 shows a cube of edge length L centered on a long thin wire with linear charge density λ. The flux through one face of the cube is not simply EA because, in this case, the electric field varies in both strength and direction. But you can calculate the flux by actually doing the flux integral. (a) Consider the face parallel to the yz-plane. Define area dA (→ above A) as a strip of width dy and height L with the vector pointing in the 𝓍-direction. One such strip is located at position y. Use the known electric field of a wire to calculate the electric flux dΦ through this little area. Your expression should be written in terms of y, which is a variable, and various constants. It should not explicitly contain any angles.