All questions
Question 1
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 +q0 from one point on the surface of the sphere to another point on the surface?
- Zero, because the electric field is zero everywhere on the surface of the sphere.
- Zero, because the electric potential is constant everywhere on the surface of the sphere. (correct answer)
- 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.
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 q0 between two points is W=q0ΔV. Since the initial and final points are on the surface, ΔV=0, and thus the work done is zero. Question 2
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 always parallel to the surface at that point.
- The electric field vector is always perpendicular to the surface at that point. (correct answer)
- 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.
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.
Question 3
Consider a system of two concentric, thin conducting spherical shells. The inner shell has radius R1 and charge +Q1. The outer shell has radius R2 and charge +Q2. The system is in electrostatic equilibrium.
What is the electric potential of the inner shell at radius R1? Assume the potential is zero at an infinite distance.
- V=4πϵ01R1Q1
- V=4πϵ01(R1Q1+Q2)
- V=4πϵ01(R1Q1+R2Q2) (correct answer)
- V=4πϵ01(R2Q1+Q2)
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=R1), the potential due to its own charge Q1 is kQ1/R1. For the outer shell, the potential inside it (for r<R2) is constant and equal to its surface potential, kQ2/R2. Thus, the total potential at R1 is the sum of these two potentials. Question 4
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 charge.
- Both spheres have the same final surface charge density.
- Both spheres have the same final electric potential. (correct answer)
- 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.
Question 5
A point charge +q is placed inside the cavity of a hollow, neutral conducting sphere, but it is not at the center. The system is in electrostatic equilibrium.
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. (correct answer)
- 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 farthest from the charge +q.
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. Question 6
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 VC, the potential at the surface be VS, and the potential at a point r=R/2 from the center be VP. Which statement correctly compares these potentials?
- VC>VP>VS
- VS>VP>VC
- VC=VP=VS (correct answer)
- VC=0, but VP and VS 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. Question 7
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 highest on the side where positive charge accumulates.
- 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 entire sphere is at a single, constant electric potential. (correct answer)
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.
Question 8
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 (correct answer)
- E=4πϵ01R2Q
- E=4πϵ01(R/2)2Q
- E=4πϵ01R3Q(R/2)
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. Question 9
An isolated, charged conducting object has a shape with one end being sharply pointed and the other end being rounded. When in electrostatic equilibrium, how does the surface charge density σ at the pointed end compare to the surface charge density at the rounded end?
- The surface charge density is greatest at the pointed end. (correct answer)
- The surface charge density is greatest at the rounded end.
- The surface charge density is uniform over the entire surface.
- The surface charge density is zero at both the pointed and rounded ends.
Explanation: For a conductor to be an equipotential, charges accumulate at points of smaller radius of curvature. A sharply pointed end has a very small radius of curvature, leading to a high concentration of charge and thus a greater surface charge density compared to the flatter, rounded end.
Question 10
A point charge +q is placed at the center of a thick, uncharged, conducting spherical shell with inner radius R1 and outer radius R2. The system is in electrostatic equilibrium.
What are the net charges induced on the inner surface (at R1) and outer surface (at R2) of the conducting shell?
- Inner surface: −q; Outer surface: +q (correct answer)
- Inner surface: +q; Outer surface: −q
- Inner surface: −q; Outer surface: 0
- Inner surface: 0; Outer surface: +q
Explanation: To make the electric field inside the conductor zero, a charge of −q must be induced on the inner surface to cancel the field from the point charge +q. Since the conducting shell was initially neutral, charge conservation requires that a charge of +q must then appear on the outer surface. Question 11
A neutral, solid conducting sphere is placed in a uniform external electric field that points to the right. After electrostatic equilibrium is reached, which statement best describes the distribution of charge on the sphere's surface?
- The sphere's surface remains electrically neutral everywhere, as no charge was added.
- The right side of the sphere becomes negatively charged and the left side becomes positively charged.
- The left side of the sphere becomes negatively charged and the right side becomes positively charged. (correct answer)
- The entire surface of the sphere becomes uniformly negatively charged due to the field.
Explanation: The external electric field exerts a force on the free electrons in the conductor. Since the field points to the right, the negatively charged electrons are pushed to the left. This accumulation of electrons makes the left side negatively charged and leaves a deficit of electrons on the right side, making it positively charged. This process is called polarization.
Question 12
A solid conducting sphere and a solid insulating sphere of the same size are each given the same amount of excess positive charge. In electrostatic equilibrium, how does the charge distribution differ between the two spheres?
- In both spheres, the excess charge is distributed uniformly throughout the volume.
- In the conductor, the charge is on the surface; in the insulator, it is distributed throughout the volume. (correct answer)
- In the conductor, the charge is distributed throughout the volume; in the insulator, it is on the surface.
- In both spheres, the excess charge resides only on the surface.
Explanation: In a conductor, charges are free to move and will repel each other to the surface to achieve equilibrium. In a non-conducting insulator, charges are not free to move and will remain distributed throughout the volume where they were placed.
Question 13
A point charge +q is placed at the center of a hollow, electrically neutral, conducting spherical shell.
What is the net electrostatic force exerted on the point charge +q?
- A net force directed radially outward, exerted by the induced negative charge on the inner surface.
- A net force directed radially inward, exerted by the induced positive charge on the outer surface.
- Zero, because the induced charges on the shell's inner surface are distributed symmetrically. (correct answer)
- Non-zero, because the induced positive charge on the outer surface attracts the inner induced negative charge.
Explanation: Due to the spherical symmetry of the shell and the central position of the charge +q, the induced negative charge on the inner surface is distributed uniformly. By symmetry, the vector sum of the forces exerted by all parts of this induced charge on the central charge is zero. Question 14
A very large, flat conducting plate has a uniform surface charge density +σ on one of its faces. The plate is in electrostatic equilibrium. What is the magnitude of the electric field at a point just outside the surface of the plate?
- E=2ϵ0σ
- E=ϵ0σ (correct answer)
- E=0
- E=ϵ02σ
Explanation: Using a Gaussian pillbox with one face inside the conductor (where E=0) and the other outside, Gauss's law ∮E⋅dA=Qenc/ϵ0 becomes EA=(σA)/ϵ0. This simplifies to E=σ/ϵ0. The field from an insulating sheet is half this value because it radiates in both directions, whereas for a conductor, the field is zero on one side. Question 15
An irregularly shaped, isolated conductor carries a net positive charge and is in electrostatic equilibrium. Point A is on a sharply curved part of the surface, and Point B is on a flatter part of the surface. How does the electric potential at Point A, VA, compare to the electric potential at Point B, VB?
- VA>VB because the charge density is higher at Point A.
- VA<VB because the radius of curvature is smaller at Point A.
- VA=VB because the entire conductor is an equipotential surface. (correct answer)
- The relationship cannot be determined without knowing the total charge.
Explanation: Regardless of the shape of the conductor or the distribution of charge on its surface, a fundamental property of a conductor in electrostatic equilibrium is that its entire volume and surface are at the same electric potential. Therefore, the potential at any two points on the surface must be equal.
Question 16
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. (correct answer)
- 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.
Question 17
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 uniformly distributed throughout the volume of the sphere.
- The excess charge is concentrated at the geometric center of the sphere.
- The excess charge is distributed entirely on the outer surface of the sphere. (correct answer)
- The excess charge is confined to a thin layer just beneath the outer surface 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.
Question 18
A sensitive electronic component must be protected from stray external static electric fields. What is the most effective method for shielding it?
- Place the component inside a solid block of a good insulating material such as glass.
- Enclose the component completely within a thin, sealed box made of plastic.
- Surround the component with a cage made of a strong insulating material like porcelain.
- Enclose the component completely within a closed box made of a conducting material. (correct answer)
Explanation: This method is known as electrostatic shielding or using a Faraday cage. When a conducting enclosure is placed in an external electric field, the free charges within the conductor redistribute on its surface to create an internal electric field that perfectly cancels the external field inside. This leaves the interior region free of electric fields.
Question 19
An isolated, neutral, solid conductor is suddenly placed in an external electric field. What is the primary cause for the redistribution of free charges that leads to electrostatic equilibrium?
- The free charges move to minimize the total electric potential energy of the conducting object.
- The charges move in response to the net electric field inside the conductor until that field becomes zero. (correct answer)
- The external field induces a magnetic field inside the conductor, which moves the charges.
- The charges are repelled by the conductor's surface and move towards the interior of the material.
Explanation: Initially, the external field penetrates the conductor, exerting a force on the free charges. These charges move, creating an induced electric field that opposes the external field. This movement continues until the induced field exactly cancels the external field everywhere inside the conductor. At that point, the net field inside is zero, the force on free charges is zero, and equilibrium is reached.
Question 20
A hollow, closed, conducting box is placed in a uniform external electric field. A small hole is then cut into one side of the box. How does this affect the electric field inside the box?
- The electric field inside remains exactly zero everywhere as long as the hole is small.
- A small, non-zero electric field may now exist inside the box, particularly near the hole. (correct answer)
- A uniform electric field, equal in magnitude to the external field, is created inside the box.
- The electric field inside becomes non-zero, but is strongest at the point farthest from the hole.
Explanation: A perfect conducting enclosure provides perfect shielding. However, introducing a hole compromises the integrity of the shield. The external field lines can now penetrate through the opening, creating a non-zero, though typically weak, electric field inside the cavity, especially in the vicinity of the hole.