Problem 12 - Master Resource Book in JEE Main Physics

Question

The electric strength of air is \(2 \times 10^{7} \mathrm{NC}^{-1}\). The maximum charge that a metaleic sphere of diameter \(6 \mathrm{~mm}\) can hold is (a) \(3 \mathrm{nC}\) (b) \(20 \mathrm{nC}\) (c) \(1.5 \mathrm{nC}\) (d) \(2 \mathrm{nC}\)

Step-by-Step Solution

Answer

The maximum charge the sphere can hold is 2 nC, so option (d) is correct.

Step 1: Understand the Problem

We need to find the maximum charge a metal sphere can hold before air breaks down and loses its insulating properties. The electric field on the surface of a sphere is connected to the charge and radius of the sphere.

Step 2: Identify Given Values

We know the electric strength of air, which is the maximum electric field air can sustain, is given as \(2 \times 10^{7} \mathrm{NC}^{-1}\). The diameter of the sphere is \(6 \mathrm{~mm}\), so the radius \(r\) is \(3 \mathrm{~mm}\) or \(3 \times 10^{-3} \mathrm{~m}\).

Step 3: Relate Electric Field and Charge

The electric field \(E\) on the surface of a sphere is given by \(E = \frac{kQ}{r^2}\), where \(k\) is the Coulomb's constant \(8.99 \times 10^9 \mathrm{Nm}^2/ ext{C}^2\), \(Q\) is the charge, and \(r\) is the radius.

Step 4: Set the Electric Field Equal to Air Strength

To find the maximum charge before breakdown, set \(E = 2 \times 10^{7} \mathrm{NC}^{-1}\). Thus, \(\frac{kQ}{r^2} = 2 \times 10^{7}\).

Step 5: Solve For Charge \(Q\)

Rearrange the equation \(Q = \frac{E \cdot r^2}{k}\). Substitute the values to compute \(Q\): \[Q = \frac{2 \times 10^{7} \cdot (3 \times 10^{-3})^2}{8.99 \times 10^9}\] \[Q = \frac{2 \times 10^{7} \cdot 9 \times 10^{-6}}{8.99 \times 10^9}\] \[Q = \frac{1.8 \times 10^2}{8.99 \times 10^9}\] \[Q \approx 2 \times 10^{-9} \mathrm{C} = 2 \mathrm{nC}\]

Step 6: Choose the Correct Option

The calculated maximum charge \(Q\) is \(2 \mathrm{nC}\). Therefore, the correct answer is option (d) \(2 \mathrm{nC}\).

Key Concepts

electric strength of air

The electric strength of air is a crucial concept in understanding how charges behave in our environment. It represents the maximum electric field that air can withstand before it starts conducting electricity.
When this threshold is exceeded, air loses its role as an insulator and becomes a conductor.

• The electric strength of air is typically around \(2 \times 10^{7} \mathrm{NC}^{-1}\).
• This value indicates the point at which ionization can occur, leading to phenomena such as sparks or lightning.

Understanding this helps in applications such as designing high-voltage equipment and ensuring electrical safety.
For instance, it is why overhead wires are placed at safe distances apart to prevent unintended discharge of electricity through the air.

charge on metal sphere

The charge on a metal sphere is influenced by several factors, including its size and the surrounding electric field.
When we talk about finding the charge a sphere can hold, it's about determining the limit before unwanted electric discharge occurs.

• The geometry of a sphere means that charge uniformly distributes over its surface.
• The radius of the sphere directly impacts the electric field on its surface.

For a given electric field strength, a larger sphere can generally hold more charge than a smaller one, because the total surface where the charge distributes is greater.
Therefore, understanding the sphere's properties and how they relate to electric fields can help calculate how much charge it can sustain safely.

Coulomb's law

Coulomb's Law provides the mathematical framework to understand the electric forces between charges.
It states that the electric force \(F\) between two point charges, \(Q_1\) and \(Q_2\), separated by a distance \(r\), is proportional to the product of their charges and inversely proportional to the square of the distance between them.
\[ F = k \frac{Q_1 Q_2}{r^2} \] Here, \(k\) is Coulomb's constant, approximately \(8.99 \times 10^9 \mathrm{Nm}^2/\mathrm{C}^2\).

• This law is fundamental in electrostatics, describing how charged objects interact in space.
• It demonstrates that as the distance between charges increases, the force decreases significantly.

The application of Coulomb's law is essential in solving problems related to electric fields, as it was in determining the maximum charge a sphere can hold before breakdown occurs, providing a practical insight into electrostatic phenomena.