(II) Two point charges, 3.4 μC and -2.0 μC, are placed 8.0 cm apart on the x axis. At what points along the x axis are the potential zero? Let V = 0 at r = ∞.

Inside a high-voltage lab, engineers have designed a storage container for electrical energy using a nonconducting sphere of radius r₂ that contains a concentric spherical cavity of radius r₁. The material between r₁ and r₂ carries a uniform charge density ρ_E ( C/m³). Determine the electric potential V, relative to V = 0 at r = ∞, as a function of the distance r from the center for r > r₂.

In a photocell, ultraviolet (UV) light provides enough energy to some electrons in barium metal to eject them from the surface at high speed. To measure the maximum energy of the electrons, another plate above the barium surface is kept at a negative enough potential that the emitted electrons are slowed down and stopped, and return to the barium surface. See Fig. 23–52. If the plate voltage is -3.02 V (compared to the barium) when the fastest electrons are stopped, what was the speed of these electrons when they were emitted?

A Van de Graaff generator (Fig. 23–58) can develop a very large potential difference, even millions of volts. Electrons are pulled off the belt by the high voltage pointed electrode (positive) at A, leaving the belt positively charged. (Recall Example 23–5 where we saw that near sharp points the electric field is high and ionization can occur.) The belt carries the positive charge up inside the spherical shell where electrons from the large conducting sphere are attracted over to the pointed conductor at B, leaving the outer surface of the conducting sphere positively charged. As more charge is brought up, the sphere reaches extremely high voltage. Consider a Van de Graaff generator with a sphere of radius 0.20 m. What is the electric potential on the surface of the sphere when electrical breakdown occurs ( E = 3 x 10⁶ V/m) ? Assume V = 0 at r = ∞.

(II) Point a is 26 cm north of a -3.8 μC point charge, and point b is 36 cm west of the charge (Fig. 23–43). Determine ${\vec{E}}_b-{\vec{E}}_a$ (magnitude and direction).

(III) Suppose the flat circular disk of Fig. 23–15 (Example 23–10) has a nonuniform surface charge density that depends on the distance R from the center of the disk: σ = aR². Find the potential V (x) at points along the x axis, relative to V = 0 at x = ∞.

(III) The dipole moment, considered as a vector, points from the negative to the positive charge. The water molecule, Fig. 23–47, has a dipole moment $\overrightarrow{p}$ which can be considered as the vector sum of the two dipole moments ${\overrightarrow{p}}_1$ and ${\overrightarrow{p}}_2$ as shown. The distance between each H and the O is about 0.96 x 10^-10 m. The lines joining the center of the O atom with each H atom make an angle of 104° as shown, and the net dipole moment has been measured to be p = 6.1 x 10^-30 C m. (a) Determine the effective charge q on each H atom. (b) Determine the electric potential, far from the molecule, due to each dipole, ${\overrightarrow{p}}_1$ and ${\overrightarrow{p}}_2$, and show that they sum to

V = (1 / 4π∊₀) (p cos θ / r²),

where p is the magnitude of the net dipole moment, $\overrightarrow{p}={\overrightarrow{p}}_1+{\overrightarrow{p}}_2$, and V is the total potential due to both ${\overrightarrow{p}}_1$ and ${\overrightarrow{p}}_2$. Take V = 0 at r = ∞.

(III) The dipole moment, considered as a vector, points from the negative to the positive charge. The water molecule, Fig. 23–47, has a dipole moment $\overrightarrow{\mathbf{p}}$ which can be considered as the vector sum of the two dipole moments $\overrightarrow{\mathbf{p}}_1$ and $\overrightarrow{\mathbf{p}}_2$ as shown. The distance between each H and the O is about 0.96 x 10^-10 m. The lines joining the center of the O atom with each H atom make an angle of 104° as shown, and the net dipole moment has been measured to be p = 6.1 x 10^-30 C m. Determine the effective charge q on each H atom.

A Van de Graaff generator (Fig. 23–58) can develop a very large potential difference, even millions of volts. Electrons are pulled off the belt by the high voltage pointed electrode (positive) at A, leaving the belt positively charged. (Recall Example 23–5 where we saw that near sharp points the electric field is high and ionization can occur.) The belt carries the positive charge up inside the spherical shell where electrons from the large conducting sphere are attracted over to the pointed conductor at B, leaving the outer surface of the conducting sphere positively charged. As more charge is brought up, the sphere reaches extremely high voltage. Consider a Van de Graaff generator with a sphere of radius 0.20 m.

(b) What is the charge on the sphere for the potential found in part (a)