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 cube has sides of $4.0\mathrm{cm}$. Inside the cube is a $20\mathrm{nC}$ and a $-30\mathrm{nC}$ charge. What is the net flux through the cube?

A $4.0\mathrm{cm}\times 5.0\mathrm{cm}$ rectangle lies in the x-y plane; let the normal vector point in the z direction. What is the flux through the rectangle of a uniform electric field $\stackrel{\rightharpoonup }{E}=(-7.8\times {10}^5\widehat{j}+2.3\times 10^5\widehat{k})N/C$

A hemispherical surface with radius r in a region of uniform electric field E→ has its axis aligned parallel to the direction of the field. Calculate the flux through the surface.

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. (c) For what angle φ between the normal to the sheet and the electric field is the magnitude of the flux through the sheet (i) largest and (ii) smallest? Explain your answers.

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?

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