24. Electric Force & Field; Gauss' Law

A hemispherical surface with radius $r$ in a region of uniform electric field $\overrightarrow{E}$ 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. For what angle $\varphi$ 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 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?

The cube in FIGURE EX24.7 contains negative charge. The electric field is constant over each face of the cube. Does the missing electric field vector on the front face point in or out? What strength must this field exceed?

A 2.0 cm × 3.0 cm rectangle lies in the $xz$-plane with unit vector $\hat{n}$ pointing in the +y-direction. What is the electric flux through the rectangle if the electric field is $\overrightarrow{E}=(4000\hat{i}-2000\hat{k})$ 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 $\hat{n}$ points in the +𝒵-direction. What is the electric flux through the rectangle if the electric field is $E=\left(2000m^{-1}\right)x\hat{k}$ 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?

Find the electric fluxes Φ_A to Φ_E through surfaces A to E in FIGURE P24.29.

A spherically symmetric charge distribution produces the electric field $\overrightarrow{E}=\left(5000r^2\right)\hat{r}$ N/C, where r is in m. How much charge is inside this 40-cm-diameter spherical surface?

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 ${\Phi }_e=Q_{in}/{ϵ}_0$ 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. Show that the net flux through the cube is ${\Phi }_e=Q_{in}/{ϵ}_0$.

FIGURE P31.38 shows the electric field inside a cylinder of radius $R=3.0$ mm. The field strength is increasing with time as $E=1.0\times {10}^8{t}^2$ 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 . Find an expression for the electric flux ${\Phi }_e$ through the entire cylinder as a function of time.