If a hollow spherical conductor is charged with positive charge the potential inside it will be

Gauss Problem: A hollow uncharged spherical conducting shell has an inner radius a and an outer radius b. A positive point charge q is in the cavity at the center of the sphere. (a) Find the charge on each surface of the conductor (surface a and surface b). (b) Find the electric field everywhere. (c) Find the potential everywhere, assuming that V = 0 at infinity. Solution: • Concepts: Gauss' law, properties of conductors • Reasoning: The field due to a spherically symmetric charge distribution can be found from Gauss' law. The inside of a conductor is field free in electrostatics. • Details of the calculation: (a) Charge on surface a: -q Charge on surface b:q (b) 0 b: E = kq/r 2 ( r/r) (c) r > b: V = kq/r a a, and E(r) = ρr/(3ε 0) for r < a. Here ρ = 3Q/4πa 3. Therefore E(r) = Qr/(4πa 3ε 0) for r < a. At r = 0 we have E(r) = 0. (b) U = (4πε 0/2)[ ∫ 0 ar 2dr Q 2r 2/(4πa 3ε 0) 2 + ∫ a ∞r 2dr Q 2/(4πε 0r 2) 2] = (Q 2/(8πε 0))[ ∫ 0 ar 4dr/a 6 + ∫ a ∞dr /r 2] = [Q 2/(8πε 0)][1/(5a) + 1/a] = (3/5) Q 2/(4πε 0a). (c) W = q[∫ 0 adr Qr/(4πa 3ε 0) + ∫ a ∞dr Q/(4πε 0r 2)] = [qQ/(4πε 0)][1/(2a) + 1/a] = 3qQ/(8πε 0 a).

There are two ways of answering your question. The first is that potential is defined up to an arbitrary constant, so you can define it to be any constant value inside the shell. The second way assumes that you mean the potential is zero at infinity. In that case: You haven't said anything about the charge outside of the shell. The potential inside will be constant, but will be equal to the potential at the surface of the shell. That potential will have a nonzero value due to the charges outside. Potential is a result of the addition of potential due to all the small area elements on the sphere. Nett Electric Field cannot be used to calculate potential. The case is analogous to the gravitational potential inside a hollow spherical shell. The gravitational potential inside the shell is constant even though the field is zero.

Let us consider a conductor with a cavity or a hole. Electric field inside the cavity is zero. Let E is the electric field just outside the conductor, q is the electric charge, σ is the charge density, and ∈ 0 ​ is the permittivity of free space. charge ∣ q ∣ = σ × d s According to Gauss's law, Flux, ϕ = E . d s = ∈ 0 ​ ∣ q ∣ ​ E d s = ∈ 0 ​ σ × d s ​ ∴ E = ∈ 0 ​ σ ​ n ^ Therefore, the electric field just outside the conductors is ∈ 0 ​ σ ​ n ^. This field is a superposition of field due to the cavity ( E ′ ) a nd the field due to the rest of the charged conductor ( E ′ ) . These fields are equal and opposite inside the conductor, and equal in magnitude and direction outside the conductor. ∴ E ′ + E ′ = E E ′ = 2 E ​ E ′ = 2 ∈ 0 ​ σ ​ n ^ Therefore, the field due to the rest of the conductor is ∈ 0 ​ σ ​ n ^. Hence, proved. $$(E_2\,-\, E_1)\, \dot\, \hat$$ where n ^ is a unit vector normal to the surface at a point and σ is the surface charge density at that point. (The direction of n ^ is from side 1 to side 2.) Hence show that just outside a conductor, the electric field is σ n ^ / ϵ 0 ​ (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.]

https://phys.libretexts.org/@app/auth/3/login?returnto=https%3A%2F%2Fphys.libretexts.org%2FBookshelves%2FUniversity_Physics%2FBook%253A_University_Physics_(OpenStax)%2FBook%253A_University_Physics_II_-_Thermodynamics_Electricity_and_Magnetism_(OpenStax)%2F06%253A_Gauss's_Law%2F6.04%253A_Applying_Gausss_Law \( \newcommand\) • • • • • • • • • • • • • • • • • • • • • • Learning Objectives By the end of this section, you will be able to: • Explain what spherical, cylindrical, and planar symmetry are • Recognize whether or not a given system possesses one of these symmetries • Apply Gauss’s law to determine the electric field of a system with one of these symmetries Gauss’s law is very helpful in determining expressions for the electric field, even though the law is not directly about the electric field; it is about the electric flux. It turns out that in situations that have certain symmetries (spherical, cylindrical, or planar) in the charge distribution, we can deduce the electric field based on knowledge of the electric flux. In these systems, we can find a Gaussian surface S over which the electric field has constant magnitude. Furthermore, if \(\vec\). Here is a summary of the steps we will follow: Problem-Solving Strategy: Gauss’s Law • Identify the spatial symmetry of the charge distribution. This is an important first step that allows us to choose the appropriate Gaussian surface. As examples, an isolated point charge has spherical symmetry, and an infinite line of cha...

11 Magnetic Forces and Fields • Introduction • 11.1 Magnetism and Its Historical Discoveries • 11.2 Magnetic Fields and Lines • 11.3 Motion of a Charged Particle in a Magnetic Field • 11.4 Magnetic Force on a Current-Carrying Conductor • 11.5 Force and Torque on a Current Loop • 11.6 The Hall Effect • 11.7 Applications of Magnetic Forces and Fields • Learning Objectives By the end of this section, you will be able to: • Calculate the potential due to a point charge • Calculate the potential of a system of multiple point charges • Describe an electric dipole • Define dipole moment • Calculate the potential of a continuous charge distribution Point charges, such as electrons, are among the fundamental building blocks of matter. Furthermore, spherical charge distributions (such as charge on a metal sphere) create external electric fields exactly like a point charge. The electric potential due to a point charge is, thus, a case we need to consider. We can use calculus to find the work needed to move a test charge q from a large distance away to a distance of r from a point charge q. Noting the connection between work and potential W = − q Δ V , W = − q Δ V , as in the last section, we can obtain the following result. E = F q t = k q r 2 . E = F q t = k q r 2 . Recall that the electric potential difference V is a scalar and has no direction, whereas the electric field E → E → is a vector. To find the voltage due to a combination of point charges, given zero voltage at...