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A solid conducting sphere having a charge \(Q\) is surrounded by an uncharged concentric conducting hollow spherical shell. The potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell is \(V\). If the shell is now given a charge of \(-3 Q\), the new potential difference between the same two surfaces is (IIT-JEE, 1989) a. \(V\) b. \(2 V\) c. \(4 \mathrm{~V}\) d. \(-2 V\)

Short Answer

Expert verified
The new potential difference is \(-2V\).

Step by step solution

01

Understanding Initial Setup

The solid conducting sphere has a charge of \(Q\), causing the electric potential at its surface to be due to this charge. The hollow spherical shell is initially uncharged, meaning no electric field exists in its interior. The potential difference \(V\) between the solid sphere surface and the outer surface of the shell is given, solely due to the charge on the sphere.
02

Formula for Potential Difference

The potential difference \(V\) can be expressed in relation to the charge \(Q\) and the radii of the spheres. However, specifics are not required as they will cancel out. Initially, \(V = k\frac{Q}{R_1} - k\frac{Q}{R_2}\), where \(k\) is Coulomb's constant, \(R_1\) the radius of the solid sphere, and \(R_2\) the radius of the hollow sphere.
03

Introducing Charge on the Shell

When the shell is given a charge of \(-3Q\), it will affect the potential on the outer surface of the shell but not within the shell itself, as the hollow region will still have no electric field.
04

Find New Potential Difference

The potential difference after adding charge to the shell will be \(V' = k\frac{Q}{R_1} - k\frac{-3Q}{R_2}\). This simplifies the expression to a new potential difference: \(V' = V + k\frac{4Q}{R_2}\).
05

Comparing to Initial Potential Difference

Assume \(V = k\frac{Q}{R_1} - k\frac{Q}{R_2}\) simplifies to \(V = V_1 - V_2\). By comparing terms, the change due to the added charge on the shell leads to a new difference resulting in a total change of \(-2V\) across the surfaces, given symmetry in charge distribution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conducting Sphere
A conducting sphere is a solid object that can distribute electric charge across its surface evenly. This property arises because the sphere is made from a conductive material, allowing electrons to move freely. When a conducting sphere is charged, the charge resides entirely on the surface. This is due to the repulsion between like charges: electrons spread as far apart as possible, which results in a uniform distribution over the sphere's surface.

The presence of a charge on a conducting sphere influences the electric potential surrounding it. An important characteristic of a conducting sphere is that inside the sphere, the electric field is zero. This means any charge placed inside does not experience a force due to the sphere's surface charge.
  • Electric field inside a conducting sphere: zero
  • Charge distribution: only on the surface
  • Influences electric potential in the surrounding space
Potential Difference
Potential difference, often called voltage, is a measure of the work done in moving a unit charge between two points in an electric field. In the context of a conducting sphere and a concentric shell, the potential difference is the work needed to move charge from the sphere's surface to the shell's surface. This is crucial in understanding how charges affect one another at a distance.

In our specific problem, the potential difference between the surface of the conducting sphere and the hollow spherical shell initially depends only on the charge of the sphere, given the shell begins with no charge. When this changes and the shell gains a charge of \( -3Q \), the potential difference between these two surfaces changes as well. Mathematically, this can be expressed as \(V = k\frac{Q}{R_1} - k\frac{Q}{R_2}\), where \(k\) is Coulomb's constant, and \(R_1\) and \(R_2\) are radii of the inner and outer spheres, respectively.
  • Definition: Work to move charge between two points
  • Depends on charge magnitude and configuration
  • Influences by adding/removing charge
Coulomb's Law
Coulomb's Law describes the force between two charged objects. It states that the force \(F\) between two point charges is directly proportional to the product of the charges (\(q_1\) and \(q_2\)) and inversely proportional to the square of the distance \(r\) between them. The equation is given by \(F = k\frac{q_1 q_2}{r^2}\), where \(k\) is the electrostatic constant.

This law is fundamental in calculating the interactions between charged objects, like in our exercise involving a conducting sphere and a spherical shell. Coulomb's Law helps determine the electric potential difference because the potential difference is derived from the electric field, which originates from these forces. Understanding how charges interact, distribute themselves, and influence potential is grounded in this foundational principle.
  • Force depends on product of charges
  • Force is inversely related to distance squared
  • Key law in electrostatics

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Most popular questions from this chapter

A large insulated sphere of radius \(r\) charged with \(Q\) units of electricity is placed in contact with a small insulated uncharged sphere of radius \(r^{\prime}\) and is then separated. The charge on the smaller sphere will now be a. \(\frac{Q\left(r^{\prime}+r\right)}{r^{\prime}}\) b. \(\frac{Q\left(r^{\prime}+r\right)}{r}\) c. \(\frac{Q r}{r^{\prime}+r}\) d. \(\frac{Q r^{\prime}}{r^{\prime}+r}\)

When a \(2 \mu C\) of charge is carried from a point \(A\) to point \(B\), the amount of work done by electric field is \(50 \mu \mathrm{J}\). What is the potential difference and which point is at a higher potential? a. \(25 \mathrm{~V}, B\) b. \(25 \mathrm{~V}, A\) c. \(20 \mathrm{~V}, B\) d. Both are at same potential

A particle of mass \(m\) carrying charge ' \(q^{\prime}\) is projected with velocity ' \(v^{\prime}\) from point ' \(P\) ' towards an infinite line of charge from a distance ' \(a\) '. Its speed reduces to zero momentarily at point \(Q\) which is at a distance \(a / 2\) from the line of charge. If anóther particle with mass \(m\) and charge \(-q^{\prime}\) is projected with the same velocity ' \(v^{\prime}\) from \(P\) towards the line of charge, what will be its speed at \(Q\) ? a. \(\sqrt{2} v\) b. \(\sqrt{3} v\) c. \(\frac{v}{\sqrt{2}}\) d. \(\frac{v}{\sqrt{3}}\)

Mark the correct statement a. An electron and a proton when released at rest in a uniform electric field experience the same force and the same acceleration. b. Two equipotential surfaces may intersect. c. A solid conducting sphere holds more charge than a hollow conducting sphere of the same radius. d. No work is done in taking a positive charge from one point to another inside a negatively charged metallic phere

Four identical charges are placed at the points \((1,0,0)\), \((0,1,0),(-1,0,0)\) and \((0,-1,0)\). Then, a. the potential at the origin is zero b. the electric field at the origin is not zero c. the potential at all points on the \(z\) -axis, other than the origin, is zero d. the field at all points on the \(z\) -axis, other than the origin, acts along the \(z\) -axis

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