Problem 179 - A Complete Resource Book in Physics for JEE Main

Question

A uniformly charged solid sphere of radius \(R\) has potential \(V_{0}\) (measured with respect to \(\infty\) ) on its surface. For this sphere, the equipotential surfaces with potential \(\frac{3 V_{0}}{2}, \frac{5 V_{0}}{4}, \frac{3 V_{0}}{4}\) and \(\frac{V_{0}}{4}\) have radius \(R_{1}, R_{2}\), \(R_{3}\), and \(R_{4}\), respectively. Then (A) \(R_{1} \neq 0\) and \(\left(R_{2}-R_{1}\right)>\left(R_{4}-R_{3}\right)\) (B) \(R_{1}=0\) and \(R_{2}<\left(R_{4}-R_{3}\right)\) (C) \(2 R\left(R_{4}-R_{3}\right)\)

Step-by-Step Solution

Answer

The correct answer is option (B), where R₁ = 0 and R₂ < (R₄ - R₃), as 0 < \(\frac{R}{3}\).

Step 1: Potential inside a uniformly charged sphere

The potential V inside a uniformly charged sphere can be expressed as: V = \(\frac{1}{4π\epsilon_0}\) \(\frac{QR^2}{(2R-r)r}\), where Q = total charge on the sphere, R = radius of the sphere, r = distance from the center of the sphere, and ε₀ = vacuum permittivity. On the sphere's surface (r = R), we have: V₀ = \(\frac{1}{4π\epsilon_0}\) \(\frac{QR^2}{(2R-R)R}\) We can solve this equation for Q and plug it into the expression for V:

Step 2: Solve for Q

Solve the V₀ equation for Q and substitute it into the expression for V: Q = \(\frac{4π\epsilon_0 V_0 R}{3}\) V(r) = \(\frac{4π\epsilon_0 V_0 R^3}{3R^2-3Rr + r^2}\) Now, we can equate V(r) to the given potential values and solve for the corresponding radii.

Step 3: Solve for radii R₁, R₂, R₃, and R₄

For each potential value, equate V(r) to it and solve for the corresponding radius: 3V₀/2 = \(\frac{4π\epsilon_0 V_0 R^3}{3R^2-3Rr_1 + r_1^2}\) => R₁ = 0 5V₀/4 = \(\frac{4π\epsilon_0 V_0 R^3}{3R^2-3Rr_2 + r_2^2}\) => R₂ = \(\frac{R}{2}\) 3V₀/4 = \(\frac{4π\epsilon_0 V_0 R^3}{3R^2-3Rr_3 + r_3^2}\) => R₃ = \(\frac{2R}{3}\) V₀/4 = \(\frac{4π\epsilon_0 V_0 R^3}{3R^2-3Rr_4 + r_4^2}\) => R₄ = R

Step 4: Verify the given options

Now, we can verify the given options one by one using our computed values for R₁, R₂, R₃, and R₄: (A) R₁ ≠ 0 and (R₂ - R₁) > (R₄ - R₃) - This is false, as R₁ = 0. (B) R₁ = 0 and R₂ < (R₄ - R₃) - This is true, as 0 < (\(\frac{R}{3}\)). (C) 2R < R₄ - This is false, as R₄ = R. (D) R₁ = 0 and R₂ > (R₄ - R₃) - This is false, as \(\frac{R}{2}\) < \(\frac{R}{3}\). So, the correct answer is option (B).

Key Concepts

Uniformly Charged Solid Sphere

Understanding the concept of a uniformly charged solid sphere is essential in comprehending various principles in electrostatics. When we refer to a solid sphere as being 'uniformly charged,' we mean that the electric charge is distributed evenly throughout its volume. Imagine a sphere made out of a non-conducting material where tiny charges are spread out so that every little volume of the sphere has the same amount of charge as any other equal-sized volume anywhere in the sphere.

This is different from a hollow sphere or a conducting sphere where the charge resides only on the surface. In a solid sphere with uniform charge, the charge doesn't just sit on the outer layer; it permeates the entire structure. For any point inside the sphere, the electric field and potential depend on the distance from the center of the sphere, which can be computed using specific formulas such as the one provided in the exercise. Once we understand this distribution, we can begin to analyze the sphere's electric potential and the surfaces around it that share the same potential, known as equipotential surfaces.

Electric Potential

Electric potential is a fundamental concept in the study of electricity and magnetism. It refers to the amount of work needed to move a unit positive charge from a reference point (usually taken at infinity) to a specific point in an electric field without producing any acceleration. It's a scalar quantity, meaning it has magnitude but no direction, and is measured in volts (V).

In simple terms, if you imagine rolling a ball up a hill, the effort you exert to push the ball to the top is analogous to the work done against the electric field to bring a charge to a point against its electric field. Inside a uniformly charged solid sphere, the electric potential decreases from the surface to the center, following a specific quadratic function of the radius, as illustrated in the solved exercise. The electric potential at any point within the sphere accounts for all the contributions from the embedded charges, and it's particularly helpful to quantify the sphere’s influence on nearby charges or other spheres.

Radius of Equipotential Surfaces

The concept of equipotential surfaces is a powerful tool in visualizing electric fields and potentials. Equipotential surfaces are imaginary shapes in a field space where each point on the surface has the same potential value. For a uniformly charged solid sphere, these surfaces are spherical and concentric with the charged sphere. The radius of an equipotential surface simply refers to how far this invisible shell is from the center of the charge distribution.

These surfaces are important because no work is done when moving a charge along an equipotential surface, as there's no change in potential energy. The problem in the textbook reveals that the potential on the surface of our solid sphere is a known value, denoted by V₀. It then inquires about the radius of surfaces where the potential is a fraction of V₀. By solving the problem step by step, we learned that those radii are not just arbitrary but correlate to specific potential values determined by the sphere's charge distribution. Understanding the radius of these surfaces helps in plotting the sphere's electric field and analyzing interactions with other charges or fields in its proximity.