Electrostatics of Conductors: Overview, Questions, Preparation

Q: Q. A spherical conducting shell of inner radius r1 and outer radius r2 has a charge Q. (a) A charge q is placed at the centre of the shell. What is the surface charge density on the inner and outer surfaces of the shell? (b) Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but has any irregular shape? Explain.

A Charge placed at the centre of the shell is +q. Hence a charge of magnitude of –q will be induced to the inner surface of the shell. Therefore, the total charge on the inner surface of the shell is –q. Surface charge density at the inner surface of the shell is given by the relation, …………(1) When a charge of magnitude of Q is placed on the outer shell and the radius of outer shell is . The total charge experienced by the outer shell will be Q + q. The surface charge density at the outer surface of the shell is given by …………(2) The electric field intensity inside a cavity is zero, even if the shell is not spherical and has any irregular shape.

Q: Q Please find the question below (a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by - Where is a unit vector normal to the surface at a point and s is the surface charge density at that point. (The direction of is from side 1 to side 2.) Hence, show that just outside a conductor, the electric field is / . (b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another. [Hint: For (a), use Gauss’s law. For, (b) use the fact that work done by electrostatic field on a closed loop is zero.]

A Electric field on one side of a charged body is and electric field on the other side of the same body is If infinite plane charged body has a uniform thickness, then electric field due to one surface of the charged body is given by, = ………….(i) Where, = unit vector normal to the surface at a point = surface charge density at that point Electric field due to the other surface of the charged body is given by = ………….(ii) due to the two surfaces, ……(iii) = = Therefore, the electric field just outside the conductor is When a charged particle is moved from one point to the other on a closed loop, the work done by the electrostatic field is zero. Hence, the tangential component of electrostatic field is continuous from one side of a charged surface to the other.

Q: Q. A long charged cylinder of linear charged density λ is surrounded by a hollow co-axial conducting cylinder. What is the electric field in the space between the two cylinders?

A Let charge density of the long charged cylinder, with length L and radius r be Another cylinder of same length surrounds the previous cylinder with radius R. Let E be the electric field produced in the space between the two cylinders. Electric flux through the Gaussian surface is given by Gaussian’s theorem as dL) = , where Then, dL) = = E = Therefore, the electric field in the space between the two cylinders is

This chapter has multiple topics listed below:

Electromagnetism Magnetic Effect of Electric Cur...Electromotive Force Curie Weiss Law Planck Equation Magnetic Force The Solenoid and the Toroid Force between two parallel curr...Moving Coil Galvanometer Torque on current loop, magneti...Potential Energy of a System of...Electrostatics of Conductors Parallel Plate Capacitor Combination of Capacitors

Dhanya B

Updated on Nov 7, 2023 17:40 IST

Conductors, like metals, contain mobile charge carriers, typically electrons. These electrons are free to move within the material, but they cannot leave the conductor. In an external electric field, they drift in the opposite direction of the field. The positive ions and bound electrons remain fixed in place.

Table of Contents

Electrostatics of Conductors
FAQs on Electrostatics of Conductors

Electrostatics of Conductors

Key results regarding the electrostatics of conductors

Inside a conductor, the electrostatic field is zero

Inside a conductor, the electric field is always zero in electrostatic equilibrium. Free charges distribute themselves to cancel out any internal electric field.

At the surface of a charged conductor, the electrostatic field must be normal to the surface at every point

At the surface of a charged conductor, the electric field is always normal (perpendicular) to the surface. This ensures that no tangential forces act on surface charges, allowing them to remain in place.

The interior of a conductor can have no excess charge in the static situation

In a static scenario, the interior of a conductor cannot have an excess charge. When a conductor is neutral, it contains equal amounts of positive and negative charges in every small volume or surface element. If the conductor is charged, any excess charge can only be found on its surface. This principle is based on Gauss's law, which states that the total electric flux through a closed surface surrounding a volume element inside a conductor is zero. Consequently, there is no net charge within the conductor, and any excess charge is limited to the surface.

The electrostatic potential is constant throughout the volume of the conductor and has the same value (as inside) on its surface

The electrostatic potential inside a conductor is constant throughout its volume and is the same as the potential on its surface. This is because the electric field is zero inside the conductor, and there is no potential difference between any points inside or on the surface. In a system of conductors with various sizes, shapes, and charge configurations, each conductor has a constant potential, which may vary from one conductor to another. If a conductor is charged, a normal electric field exists at the surface, leading to a different potential compared to a point just outside the surface.

Electrostatic shielding

Electrostatic shielding is a remarkable phenomenon observed in conductors with cavities. Regardless of the cavity's size, shape, the charge on the conductor, or external electric fields, the electric field inside the cavity is always zero. This result holds even when charges are induced on the conductor's surface by external fields, and all charges are confined to the outer surface of the conductor with a cavity. Electrostatic shielding ensures that any cavity in a conductor remains isolated from external electric influences, making it a valuable tool for protecting sensitive instruments from electrical interference.

FAQs on Electrostatics of Conductors

Q. A spherical conducting shell of inner radius r1 and outer radius r2 has a charge Q. (a) A charge q is placed at the centre of the shell. What is the surface charge density on the inner and outer surfaces of the shell? (b) Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but has any irregular shape? Explain.

Charge placed at the centre of the shell is +q. Hence a charge of magnitude of –q will be induced to the inner surface of the shell. Therefore, the total charge on the inner surface of the shell is –q.

Surface charge density at the inner surface of the shell is given by the relation,

…………(1)

When a charge of magnitude of Q is placed on the outer shell and the radius of outer shell is . The total charge experienced by the outer shell will be Q + q. The surface charge density at the outer surface of the shell is given by

…………(2)

The electric field intensity inside a cavity is zero, even if the shell is not spherical and has any irregular shape.

Q Please find the question below (a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by - Where is a unit vector normal to the surface at a point and s is the surface charge density at that point. (The direction of is from side 1 to side 2.) Hence, show that just outside a conductor, the electric field is / . (b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another. [Hint: For (a), use Gauss’s law. For, (b) use the fact that work done by electrostatic field on a closed loop is zero.]

Electric field on one side of a charged body is and electric field on the other side of the same body is If infinite plane charged body has a uniform thickness, then electric field due to one surface of the charged body is given by,

= ………….(i)

Where,

= unit vector normal to the surface at a point

= surface charge density at that point

Electric field due to the other surface of the charged body is given by

= ………….(ii)

due to the two surfaces,

……(iii)

= =

Therefore, the electric field just outside the conductor is

When a charged particle is moved from one point to the other on a closed loop, the work done by the electrostatic field is zero. Hence, the tangential component of electrostatic field is continuous from one side of a charged surface to the other.

Q. A long charged cylinder of linear charged density λ is surrounded by a hollow co-axial conducting cylinder. What is the electric field in the space between the two cylinders?

Let charge density of the long charged cylinder, with length L and radius r be

Another cylinder of same length surrounds the previous cylinder with radius R.

Let E be the electric field produced in the space between the two cylinders.

Electric flux through the Gaussian surface is given by Gaussian’s theorem as

dL) = , where

Then, dL) = =

E =

Therefore, the electric field in the space between the two cylinders is

Q. A small sphere of radius r1 and charge q1 is enclosed by a spherical shell of radius r2 and charge q2. Show that if q1 is positive, charge will necessarily flow from the sphere to the shell (when the two are connected by a wire) no matter what the charge q2 on the shell is.

A. According to Gauss’s law, the electric field between a sphere and a shell is determined by the charge on a small sphere. Hence, the potential difference, V, between the sphere and the shell is independent of charge . For positive charge , potential difference V s always positive.