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Expression For Electric Potential at any Point Due to an Electric Dipole

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ELECTRIC POTENTIAL AT ANY POINT DUE TO AN ELECTRIC DIPOLE :  Obtain expression for the electric potential at any point due to an electric dipole. Rewrite this expression if point of observation lies on the (i) axial line of the dipole and (ii) equatorial line of the dipole. Consider any point P at a distance r from the centre (O) of the electric dipole AB. Let OP make an angle $\theta$ with the dipole moment $\vec{p}$. Let $r_1$ and $r_2$ be the distances of point P from -q charge and +q charge of the dipole respectively. Step 1. Potential at point P due to -q charge is given by $V_1 = \frac{1}{4\pi\epsilon_0} \frac{(-q)}{r_1}$ Potential at point P due to +q charge is given by $V_2 = \frac{1}{4\pi\epsilon_0} \frac{q}{r_2}$ $\therefore$ Using principle of superposition, potential at point P due to the dipole is given by $V = V_1 + V_2$ or $V = -\frac{1}{4\pi\epsilon_0} \frac{q}{r_1} + \frac{1}{4\pi\epsilon_0} \frac{q}{r_2}$ $V= \frac{q}{4\pi\epsilon_0} \left[ \frac{1}{r_2} - \frac{1...

Notes : Class 12 Physics Chapter 8 Electromagnetic Waves - Physicskund

Notes : CBSE Class 12 Physics Chapter 8 Electromagnetic Waves - Physicskund 1. Basic idea of displacement current,  2. Electromagnetic waves, their characteristics, their transverse nature (qualitative idea only). 3. Electromagnetic spectrum (radio waves, microwaves, infrared, visible, ultraviolet, X-rays,gamma rays) including elementary facts about their uses.

NCERT Solution CLASS 9 SCIENCE CHAPTER 7 MOTION - Physicskund

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NCERT CLASS 9 SCIENCE CHAPTER 7 MOTION COMPLETE SOLUTION - Phyaicskund Class 9 Science Chapter 7 motion Page number 74 Question 1 to 3 Solution with video Explanation : Question 1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example. Solution : Yes, An object instead of moving through a distance can have zero displacement. Example: If an object travels from point A and reaches to the same point A, then its displacement is zero. Total Distance = 4 + 4 = 8 m Distance = Final Position - Initial Position                 = 0-0 = 0 m Question 2. A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position? Solution : Given Side of square field = 10m So, Perimeter of a square = 4 × side = 4×10 m = 40 m Farmer takes 4...

Polarisation of Light Notes Class 12 PDF | NCERT, Board, NEET & JEE

Polarisation of Light Notes Class 12 PDF | NCERT, Board, NEET & JEE Polarisation Polarisation is an important property of light that proves light is a transverse wave. It is possible only in transverse waves because their vibrations occur perpendicular to the direction of propagation. Definition: Polarisation is the phenomenon in which the vibrations of a transverse wave are restricted to one direction (or one plane) perpendicular to the direction of propagation. Important Points: Only transverse waves can be polarised. Longitudinal waves cannot be polarised. Polarisation proves the transverse nature of light. Transverse Wave A transverse wave is a wave in which the particles of the medium vibrate perpendicular to the direction of propagation. For a wave travelling along the x-axis , the particles may vibrate along the y-axis or z-axis . Remember: The direction of vibration is always perpendicular to the direction of propagation. y-Polarised Wa...