Problem 59 - Master Resource Book in JEE Main Physics

Question

A solid metallic sphere has a charge \(+3 Q .\) Concentric with this sphere is a conducting spherical shell having charge \(-Q\). The radius of the sphere is \(a\) and that of the spherical shell is \(b(b>a)\). What is the electric field at a distance \(R(a

Step-by-Step Solution

Answer

The electric field at distance R is \( \frac{3Q}{4 \pi \varepsilon_0 R^2} \) (option a).

Step 1: Understanding the Problem

We have a solid metallic sphere with a charge of +3Q and a conducting spherical shell with a charge of -Q. We need to find the electric field at a distance R from the center, where a < R < b. Given the charge configuration, we must consider the effects of both the sphere and the shell on the electric field in this region.

Step 2: Applying Gauss's Law

To find the electric field at distance R, use Gauss's Law. For a spherical Gaussian surface with radius R (where a < R < b), the net charge enclosed is the charge on the solid sphere, which is +3Q. The charge of the spherical shell does not contribute within this radius.

Step 3: Calculating the Electric Field

According to Gauss's Law, the electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space, \(\varepsilon_0\). Thus, the electric field E at distance R is given by \( E \cdot 4 \pi R^2 = \frac{3Q}{\varepsilon_0} \).

Step 4: Solving for Electric Field

Solving the equation \( E \cdot 4 \pi R^2 = \frac{3Q}{\varepsilon_0} \) for E gives us \( E = \frac{3Q}{4 \pi \varepsilon_0 R^2} \). This matches option (a).

Key Concepts

electric field

An electric field is a physical field produced by electrically charged objects. It affects other charges in the vicinity and is a vector field, meaning it has both a magnitude and direction. Generally, the strength of the electric field

• varies with distance from the charge source,
• is typically represented by vector lines pointing away from positive charges and towards negative charges,
• is defined mathematically as the electric force per unit charge,

which can be expressed by the equation:\[E = \frac{F}{q}\]where \(E\) is the electric field, \(F\) is the electric force, and \(q\) is the charge.

To find the electric field resulting from multiple charges, such as in our problem, we often use principles like superposition and Gauss's Law. Gauss's Law is particularly useful for symmetric charge distributions. It states that the electric flux through a closed surface is proportional to the enclosed charge. This simplifies calculations when dealing with geometrically simple structures like spheres.

metallic sphere

A metallic sphere is a common object in electrostatics, typically serving as a conductor. Conductors like metallic spheres allow charges to move freely across their surfaces.

Unique characteristics of metallic spheres include:

• Charge distribution: Excess charge resides on the outer surface.
• Electric field inside: There is no electric field within a conducting sphere in electrostatic equilibrium because charges distribute on the surface.
• External effects: The sphere influences the electric field environment around it.

In our exercise, the metallic sphere has a total charge of \(+3Q\). By using Gauss's Law for a Gaussian surface outside this sphere but inside the spherical shell, we can deduce that the sphere dictates the electric field in this scenario.

spherical shell

A spherical shell in electrostatics typically refers to a hollow conductor shaped like a shell, often with symmetrically curved surfaces. The behavior of charges and electric fields due to spherical shells follows certain principles:

• Charge placement: Without external charges, charges reside on the outer surface.
• Effect on internal field: The shell's own charge doesn’t contribute to the electric field inside the hollow until beyond its surface.
• Field inside the hollow: No electric field emerges within its cavity.

In our example, a conducting spherical shell is concentric with the metallic sphere. The shell has a net charge of \(-Q\), but importantly, this charge does not affect the electric field within the region between the metallic sphere and the inner surface of the shell, due to the nature of conductors.

enclosed charge

Enclosed charge refers to the total charge contained within a certain region of space. In the context of Gauss's Law, it's crucial to determine the electric field.

To find the enclosed charge, you:

• Identify your Gaussian surface.
• Calculate the total charge within this surface.
• Apply Gauss’s Law to find the relation with the electric field.

In the given problem, the Gaussian surface between the metallic sphere and the spherical shell encloses the charge of \(+3Q\) from the sphere. Since the charge \(-Q\) of the shell exerts no effect within this boundary, only \(+3Q\) influences the electric field. Hence, by applying Gauss's Law formula, we find that this enclosed charge determines the electric field strength at a given point within the specified range from the sphere's center.