Table of Contents
Charging of Insulators
Since charge cannot flow through insulators, neither conduction nor induction can be used to charge, insulators, so in order to charge an insulator friction is used. Whenever an insulator is rubbed against a body exchange of electrons takes place between the two. This results in appearance of equal and opposite charges on the insulator and the other body. Thus the insulator is charged.
For example rubbing of plastic with fur, silk with glass causes charging of these things.
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Reynolds Number
Last modified on:3 years agoReading Time:10MinutesWhat is Reynolds Number? The Reynolds number (Re) is a dimensionless parameter used in fluid mechanics to characterize the flow of a fluid (liquid or gas) around an object or through a conduit. It relates the inertial forces to the viscous forces in the fluid and helps determine the type of…
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NCERT Solution For Class 12 Physics Chapter 11 Dual Nature of radiation and matter
Last modified on:5 years agoReading Time:77MinutesQuestion 11.1: Find the (a) maximum frequency, and (b) minimum wavelength of X-rays produced by 30 kV electrons. ANSWER: Potential of the electrons, V = 30 kV = 3 × 104 V Hence, energy of the electrons, E = 3 × 104 eV Where, e = Charge on an electron = 1.6 × 10−19 C (a)Maximum frequency produced by the X-rays = ν The energy of the electrons is…
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NCERT Solutions For Class 12 Physics Chapter 8 Electromagnetic Waves
Last modified on:5 years agoReading Time:28Minutes1.(a) Figure 8.6 shows a capacitor made of two circular plates each of radius 12 cm, and separated by 5.0 cm. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 0.15A. Calculate the capacitance and the rate of change…
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NCERT Solutions for CBSE Class 12 Physics ‒ Chapter 1: Electric Charges and Fields
Last modified on:5 years agoReading Time:59MinutesQuestion1.1: What is the force between two small charged spheres having charges of 2 × 10–7 C & 3 × 10–7 C placed 30 cm apart in air? Solution1.1: Given: Q1= 2 × 10-7 C Q2 = 3 × 10-7 C r = 30 × 10-2 m We know, F = (Q1 Q2)/(4 π εo r2 ) Substituting the given values, we get…
