Skip to main content

Action at Points | Class 11 NEB Physics | Physics in Depth

-
Action at Points | Class 11 NEB Physics | Physics in Depth

Action at Points:

Action at Points Conductor

Action at points refers to the phenomenon where a charged pointed end of a conductor ionizes the surrounding air causing a continuous leakage of charge from that point.

Mechanism:

For any charged conductor, the electric charge distributes itself on the outer surface. For a pointed end, the surface charge density (\(\sigma\), charge per unit area) is inversely proportional to the radius of curvature (\(r\)) of that surface:\[\sigma \propto \frac{1}{r}\]

The electric field \(E\) is proportional to \(\sigma\). So, the two case may arise:

Case I: The electric field if immensely greater may ionize the air molecules directly stripping the electrons from it (air molecules then becomes positively charged).
Case II: The higher electric field intensity due to the pointed end may provide additional kinetic energy to the free electrons present in the atmosphere (due to radioactive phenomenon). This free electron then knocks off the electron from the air molecules. This process repeats and the electrons are attracted towards the pointed positively charged end gradually neutralizing the excess positive charges in the conductor’s surface. The ionized air molecules are however repelled. So, the wind blows away from the pointed end. It can be demonstrated by placing the candle in the right side. The burning candle tends to blow off as directed towards the direction of electric wind as in figure.
Electric Wind Candle Demonstration
Figure: Experimental demonstration of the electric wind (Common for Case I and Case II)
Special Thanks to Mr. Ashish Acharya, Kathmandu Model Secondary School.

Comments

Post a Comment

Popular Posts

Most important questions | Class 12 NEB Physics

-
Warning! The questions below are frequenty asked questions in NEB examination. Please follow the syllabus thoroughly. Every chapter and topic of your textbook is important. This questions are only for practice. Mechanics Define moment of inertia. Derive an expression for the moment of inertia of thin uniform rod about an axis through its center and perpendicular to its length. Derive a relation between the torque and angular acceleration in case of a rigid body. State the conservation of angular momentum. Derive the relation between torque and angular momentum. Write the physical significance of moment of inertia. Show that K.E. = \(\frac{1}{2} I \omega^2\) where the symbols have their usual meanings. Define simple harmonic motion. Show that the bob of a simple pendulum may move with simple harmonic motion and find its time period. Find an expression for the energy of particle in SHM...
Read more

Class 11 NEB model question solution 2077 | Physics | Complete explanation and notes

-
Image
Model Question Solution Group 'A' 1. A metre rule is used to measure the length of a piece of string in a certain experiment. It is found to be 20 cm long to the nearest millimeter. How should this result be recorded in a table of results? (a) 0.2000 m (b) 0.200 m (c) 0.20 m (d) 0.2 m Explanation: From the question, it is obvious that the least count of the metre scale is 1 mm as it measures upto the millimeter. Since, 1 mm = 0.001 m . So, result should have three significant figures. Thus, \[\begin{align*} 20 cm &= \frac{20}{100}\\ \therefore 20 cm &=0.200 \hspace{0.01cm} m\\ \end{align*}\] 2. Forces are applied to a rigid body. The forces all act in the same plane. In which diagram is the body in equilibrium? (b) Explanation: In this case, the downward force has balanced the upward force and hence the body is in equilibrium. 3. An athlete makes a long jump a...
Read more

Electromagnetic induction | Class 12 NEB Physics | Numerical problem solution

-
The magnetic flux passing perpendicular to the plane of coil is given by φ = 4t 2 +5t+2 where φ is in weber and t is in second. Calculate the magnitude of instantaneous emf induced in the coil when t = 2 sec. Solution: Given, magnetic flux, φ = 4t 2 +5t+2 time, t = 2 sec induced emf, ε = ? From Faraday's law, \[\begin{align*} \epsilon& = -\frac{d\phi}{dt}\\ &=-\frac{d(4t^2+5t+2)}{dt}\\ &=-(8t+5)\\ \end{align*}\] At t = 2 sec, the induced emf is , \[\begin{align*} \epsilon&=-(8 \times 2 + 5)\\ \therefore \epsilon& = -21\hspace{0.1cm}\text{V}\\ \end{align*}\] Thus, the magnitude of induced emf is 21 V. A straight conductor of length 25 cm is moving perpendicular to its length with a uniform speed of 10 m/s making an angle of 45 0 with a uniform magnetic field of 10 T. Calcuate the emf induced across its length. Solution: Given, length, l = 25 cm = 0.25 m spe...
Read more