Problem 92 - Master Resource Book in JEE Main Physics

Question

A sphere of radius \(r\) is charged to a potential \(V\). The outward pull per unit area of its surface is given by (a) \(\frac{4 \pi \varepsilon_{0} V^{2}}{r^{2}}\) (b) \(\frac{\varepsilon_{0} V^{2}}{2 r^{2}}\) (c) \(\frac{2 \pi \varepsilon_{0} V^{2}}{r^{2}}\) (d) \(\frac{\varepsilon_{0} V^{2}}{4 r^{2}}\)

Step-by-Step Solution

Answer

The correct answer is (b) \(\frac{\varepsilon_0 V^2}{2 r^2}\).

Step 1: Understand the Problem

We need to find the electric outward pull per unit area for a charged sphere given its radius \(r\) and potential \(V\). This is essentially the surface tension due to electric field.

Step 2: Know the Formula for Electric Pressure

The electric pressure (outward pull) on the surface of a conductor is given by \( \frac{1}{2} \varepsilon_0 E^2 \), where \( E \) is the electric field at the surface of the conductor.

Step 3: Determine the Electric Field on the Surface of the Sphere

The electric field on the surface of a sphere with surface potential \( V \) is given by \( E = \frac{V}{r} \).

Step 4: Substitute Electric Field into Electric Pressure Formula

Substitute \( E = \frac{V}{r} \) into the pressure formula: \[ P = \frac{1}{2} \varepsilon_0 \left( \frac{V}{r} \right)^2\]

Step 5: Simplify the Expression

Simplify the pressure expression:\[ P = \frac{1}{2} \varepsilon_0 \frac{V^2}{r^2}\].

Step 6: Match the Simplified Expression with Provided Options

Compare the simplified expression \( \frac{1}{2} \varepsilon_0 \frac{V^2}{r^2} \) with the choices provided: the correct match is (b) \( \frac{\varepsilon_0 V^2}{2 r^2} \).

Key Concepts

Electric Potential

Electric potential, often denoted as \( V \), is a measure of the potential energy per unit charge at a point in an electric field. It's like a landscape map indicating how much work is needed to bring a unit of positive charge from a reference point, typically infinity, to a desired location without any acceleration.

Electric potential is measured in volts (V), which corresponds to joules per coulomb. It does not depend on the path taken by the charge. Instead, it relies on the position in the electric field. When dealing with a uniformly charged sphere, the potential at the surface is given by the sphere's total charge divided by its radius. This allows us to understand how a charged object's geometric and physical characteristics influence its electric field's strength around it.

Key points:

• Electric potential is a scalar quantity.
• Its unit is volts (V).
• It helps to predict the behavior of charges in the field.
• For a sphere, its magnitude is a function of charge and radius.

Electric Field

The electric field \( E \) represents the force experienced by a unit positive charge in the vicinity of other charges. It is a vector field, which means it has both magnitude and direction. The direction of the electric field is always from positive to negative charges.

In the context of a charged sphere, the electric field at the surface is determined using the formula \( E = \frac{V}{r} \), where \( V \) is the electric potential, and \( r \) is the radius of the sphere. This formula helps you understand how the electric potential translates into an electric force acting on charges near the surface.

Important points:

• Electric field indicates the spatial influence of electric forces.
• It determines the acceleration of charges in the field.
• The magnitude of \( E \) diminishes with increased distance from the source charge (sphere in this case).
• Unit of \( E \) is volts per meter (V/m).

Surface Tension

Though traditionally associated with liquids, surface tension in electrostatics describes the mechanical stress an electric field exerts on a conductor's surface. Here, it’s the force experienced per unit area due to an electric field, similar to a force exerted along a surface in fluids.

The surface tension equation in electrostatics is essentially the electric pressure calculated from the electric field and involves components of both the sphere's charged nature and its geometry. It plays a crucial role in understanding how conductors interact with electric fields—given by the equation derived in our solution walkthrough: \( P = \frac{1}{2} \varepsilon_0 \frac{V^2}{r^2} \). This reflects how the electric potential and the sphere's size affect the outward force per unit area.

Highlights:

• Electrostatic surface tension links the pressure exerted by electric fields on surfaces.
• It is akin to mechanical stress acting on the conductor's surface.
• Calculated using the electric potential and geometry of charged objects.
• Used to predict force distribution on charged surfaces.

Electric Pressure

Electric pressure, or electrostatic pressure, is a concept that describes how an electric field exerts force on the surface of a conductor. Think of it as the push or pull experienced across a surface due to electric forces. In the given problem, electric pressure is represented by the formula \( P = \frac{1}{2} \varepsilon_0 E^2 \), where \( \varepsilon_0 \) is the permittivity of free space and \( E \) is the electric field strength at the conductor's surface.

Using our example of a sphere, this pressure becomes \( P = \frac{1}{2} \varepsilon_0 \left(\frac{V}{r}\right)^2 \). This expression shows how the electric potential \( V \), concerning the sphere's radius \( r \), defines the surface pressure exerted outward due to electrostatic forces.

Considerations:

• Electric pressure correlates with both field strength and surface charge distribution.
• Increases with rising electric field defenses at the conductor's surface.
• Critical for understanding conductor stability in electric fields.
• Helps determine potential deformation in surfaces under electric forces.