Question

If a charged spherical conductor of radius \(10 \mathrm{~cm}\) has potential \(\mathrm{v}\) at a point distant \(5 \mathrm{~cm}\) from its centre, then the potential at a point distant \(15 \mathrm{~cm}\) from the centre will be \(\ldots . .\) (A) \((1 / 3) \mathrm{V}\) (B) \((3 / 2) \mathrm{V}\) (C) \(3 \mathrm{~V}\) (D) \((2 / 3) \mathrm{V}\)

Short answer

The potential at a point distant 15 cm from the center of the charged spherical conductor is (D) \(\frac{2}{3} V\).

Step-by-step solution

Find the potential at the surface of the sphere #

To find the potential at the surface of the sphere, we recognize that the distance of the surface from the center of the sphere is equal to its radius, which is 10 cm. Thus, we can utilize the given information on the potential at 5 cm. Since the potential is inversely proportional to the distance, we can write: \[V_{surface} = V_{5cm} * \frac{10}{5} = 2 V_{5cm}\]

Find the potential at the 15 cm distance #

Now that we have the potential at the surface of the sphere, we can use it to find the potential at the point 15 cm from its center. Since the potential is inversely proportional to the distance, we can write: \[V_{15cm} = V_{surface} * \frac{10}{15} = \frac{2}{3} V_{surface}\]

Substitute the expression for V_surface from Step 1 #

Substitute the expression for V_surface found in Step 1 into the equation for V_15cm: \[V_{15cm} = \frac{2}{3} (2 V_{5cm}) = \frac{4}{3} V_{5cm}\] This corresponds to the option: (A) \((1/3)V\) (incorrect) (B) \((3/2)V\) (incorrect) (C) \(3V\) (incorrect) (D) \((2/3)V\) (correct) The potential at a point distant 15 cm from the center of the charged spherical conductor is (D) \(\frac{2}{3} V\).

Key concepts

Potential of a Sphere

The potential of a sphere refers to the electric potential on or around a spherical conductor. The formula for the potential at a point on the surface of the sphere is derived from the electric field caused by the surface charge and is given by:

• The formula: \[ V = \frac{kQ}{r} \]where \(V\) is the electric potential, \(k\) is the Coulomb's constant, \(Q\) is the charge on the sphere, and \(r\) is the radius of the sphere.
• In a charged sphere, the potential is the same everywhere on its surface. This is because the charges distribute evenly, making the field inside uniformly zero and constant outside.
• The potential outside the sphere's surface decreases with distance because it behaves like a point charge from afar.

In the exercise, the potential at any point can be determined based on its distance from the center of the sphere by assuming the sphere is a point charge when viewed from a distance, proportionally modifying the potential depending on how far you are from the surface.

Charged Conductors

Charged conductors have particular properties when it comes to electrical fields and potentials. Understanding these properties helps clarify why and how the potential on a sphere behaves as it does.

• A key property is that the excess charge on a conductor will reside entirely on its surface due to the repulsive nature of like charges.
• Inside the conductor, the electric field is zero. This is because any internal charge would cause the field to redistribute until it reaches an equilibrium.
• The potential inside and on the surface remains constant because any two points inside or on a conductor surface should be at the same potential.

For our spherical conductor, this means the potential everywhere on the surface and inside is uniform. When calculating at a point away from the surface, it's treated as if originating from a point charge at the center.

Electric Potential

Electric potential is an essential concept in electrostatics that defines the amount of work needed to move a charge from a reference point to a specific point inside the field without producing any acceleration.

• It is measured in volts (V) and is a scalar quantity.
• The potential represents electric potential energy per unit charge, meaning influences on a charge are related to the energy state within the field.
• Electric potential decreases with increasing distance from a charged object because the work required to move a charge increases.

In the given exercise, the problem uses the concept that electric potential is inversely proportional to distance from a point in space. This relationship is key in understanding why the potential is calculated as it appears, where moving away from a charged object leads to a proportional decrease in potential.