Electrostatics with Conductors
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AP Physics C: Electricity and Magnetism › Electrostatics with Conductors
A solid conducting sphere of radius $$R$$ holds a net charge $$+Q$$ and is in electrostatic equilibrium. How much work is done by an external agent to move a test charge $$+q_0$$ from one point on the surface of the sphere to another point on the surface?
Zero, because the electric potential is constant everywhere on the surface of the sphere.
A positive value, because work must be done against the repulsive electric force.
A value that depends on the path taken between the two points on the surface.
Zero, because the electric field is zero everywhere on the surface of the sphere.
Explanation
A conductor in electrostatic equilibrium is an equipotential surface, meaning the electric potential $$V$$ is the same at all points on the surface. The work done by an external agent in moving a charge $$q_0$$ between two points is $$W = q_0 \Delta V$$. Since the initial and final points are on the surface, $$\Delta V = 0$$, and thus the work done is zero.
A hollow, uncharged, conducting spherical shell is placed in a region of space with no external electric fields. What is the magnitude of the electric field inside the empty cavity of the shell?
It is zero everywhere inside the cavity.
It is non-zero and uniform, and directed radially outward from the center.
It depends on the thickness of the shell material.
It cannot be determined without knowing the dimensions of the shell.
Explanation
By Gauss's Law, if we draw a Gaussian surface inside the cavity, it encloses no charge. Since there are no charges inside, and any charges on the conductor would be on the outer surface (of which there are none since it's uncharged), the electric field inside the cavity must be zero everywhere. This is a basic case of electrostatic shielding.
A charged conductor of arbitrary shape is in electrostatic equilibrium. Which statement best describes the direction of the electric field vector at any point just outside the surface of the conductor?
The electric field vector is at a 45-degree angle to the surface at that point.
The electric field vector has no component perpendicular to the surface at that point.
The electric field vector is always perpendicular to the surface at that point.
The electric field vector is always parallel to the surface at that point.
Explanation
In electrostatic equilibrium, the surface of a conductor is an equipotential surface. If the electric field had a component parallel to the surface, it would exert a force on the charges and cause them to move along the surface, which contradicts the condition of equilibrium. Therefore, the electric field must be perpendicular to the surface.
What is the electric potential of the inner shell at radius $$R_1$$? Assume the potential is zero at an infinite distance.
$$V = \frac{1}{4\pi\epsilon_0} \frac{Q_1}{R_1}$$
$$V = \frac{1}{4\pi\epsilon_0} (\frac{Q_1+Q_2}{R_2})$$
$$V = \frac{1}{4\pi\epsilon_0} (\frac{Q_1+Q_2}{R_1})$$
$$V = \frac{1}{4\pi\epsilon_0} (\frac{Q_1}{R_1} + \frac{Q_2}{R_2})$$
Explanation
The total potential at any point is the sum of the potentials from each charged object. At the surface of the inner shell ($$r=R_1$$), the potential due to its own charge $$Q_1$$ is $$kQ_1/R_1$$. For the outer shell, the potential inside it (for $$r < R_2$$) is constant and equal to its surface potential, $$kQ_2/R_2$$. Thus, the total potential at $$R_1$$ is the sum of these two potentials.
Two isolated conducting spheres, Sphere 1 with radius $$R$$ and Sphere 2 with radius $$2R$$, are located far apart. Sphere 1 has charge $$+Q$$ and Sphere 2 is uncharged. They are then connected by a long, thin conducting wire. After the system reaches electrostatic equilibrium, which statement is correct?
Both spheres have the same final electric potential.
Both spheres have the same final surface charge density.
Both spheres have the same final charge.
All the charge transfers from Sphere 1 to the larger Sphere 2.
Explanation
When the two conducting spheres are connected by a conducting wire, they form a single conductor. In electrostatic equilibrium, the entire conductor (both spheres and the wire) must be at the same electric potential. Charge will redistribute between the spheres until this condition is met.
Which statement best describes the distribution of the induced charge on the inner surface of the conducting sphere?
The induced charge is $$-q$$ and it is distributed uniformly on the inner surface.
The induced charge is $$-q$$ and its density is highest on the part of the inner surface closest to the charge $$+q$$.
The induced charge is $$-q$$ and its density is highest on the part of the inner surface farthest from the charge $$+q$$.
The induced charge is $$+q$$ and it is distributed uniformly on the inner surface.
Explanation
An induced charge of $$-q$$ must appear on the inner surface to shield the conductor. Because the point charge $$+q$$ is off-center, it will attract the induced negative charges more strongly on the side closer to it. Therefore, the surface charge density of the induced negative charge will be non-uniform, being most concentrated on the region of the inner surface nearest to $$+q$$.
An isolated solid conducting sphere of radius $$R$$ has a net charge $$+Q$$ and is in electrostatic equilibrium. Let the electric potential at the center of the sphere be $$V_C$$, the potential at the surface be $$V_S$$, and the potential at a point $$r=R/2$$ from the center be $$V_P$$. Which statement correctly compares these potentials?
$$V_C > V_P > V_S$$
$$V_C = V_P = V_S$$
$$V_S > V_P > V_C$$
$$V_C = 0$$, but $$V_P$$ and $$V_S$$ are non-zero.
Explanation
For a conductor in electrostatic equilibrium, the electric field inside is zero. Since $$E = -dV/dr$$, a zero electric field implies that the potential $$V$$ is constant throughout the conductor. Therefore, the potential at the center, at any interior point, and at the surface are all equal.
A neutral, isolated conducting sphere is placed in a region with a uniform external electric field. After the sphere reaches electrostatic equilibrium, which statement is true regarding the electric potential of the sphere?
The potential is lowest on the side where negative charge accumulates.
The potential is zero everywhere within and on the sphere because it is neutral.
The potential is highest on the side where positive charge accumulates.
The entire sphere is at a single, constant electric potential.
Explanation
A key property of any conductor in electrostatic equilibrium, whether charged or neutral in an external field, is that it is an equipotential. The free charges redistribute themselves precisely so that the net electric field inside the conductor is zero, which means there is no potential difference between any two points within or on the conductor.
A solid, isolated conducting sphere is given a net negative charge. Once the sphere reaches electrostatic equilibrium, where is the excess charge located?
The excess charge is distributed entirely on the outer surface of the sphere.
The excess charge is concentrated at the geometric center of the sphere.
The excess charge is confined to a thin layer just beneath the outer surface of the sphere.
The excess charge is uniformly distributed throughout the volume of the sphere.
Explanation
In a conductor at electrostatic equilibrium, the free charges (electrons) repel each other and move as far apart as possible, which means they reside on the outermost surface. Additionally, the electric field inside the conductor must be zero, which can only be achieved if all net charge is on the surface.
A solid conducting sphere of radius $$R$$ is charged with a net positive charge $$+Q$$. The sphere is in electrostatic equilibrium. What is the magnitude of the electric field at a distance $$r = R/2$$ from the center of the sphere?
$$E = 0$$
$$E = \frac{1}{4\pi\epsilon_0} \frac{Q}{(R/2)^2}$$
$$E = \frac{1}{4\pi\epsilon_0} \frac{Q}{R^2}$$
$$E = \frac{1}{4\pi\epsilon_0} \frac{Q(R/2)}{R^3}$$
Explanation
A fundamental property of a conductor in electrostatic equilibrium is that the net electric field inside the material of the conductor is zero. Since the point $$r = R/2$$ is inside the conductor, the electric field must be zero.