In the 12th class physics, Chapter 1 focuses on electrostatics, introducing students to the principles and concepts related to electric charges and their interactions. Within this chapter, students encounter various short questions that require concise yet comprehensive answers to test their understanding of the topic. By answering these short questions, students can deepen their understanding of electrostatics and develop their analytical and problem-solving skills. Given below is the Short Question Answers which you can attempt for the better preparation for exams.
12th Class Physics Chapter 1 Electrostatics Short Question Answers Below
E α 1/r
E = E1 – E2 = 0
Hence the net force on the positive charge is zero.Thus it will remain at rest.
b) When a positive point charge is placed between parallel plates value of resultant electric intensity is
E = E1 + E2
because E1 due to one plate is equal in magnitude but in the same direction of E2 due to other plate.Hence net force on the charge will be from positive to negative plate.Thus it will be accelerated towards negative plate.
If two alike point charges q1 and q2 are placed at a distance r then the charge q1 exerts a force F21 on the charge q2 which acts in direction away from q1.Similarly at the same time the charge q2 exerts a force F12 on the charge q1 which is equal in magnitude to the force F21 exerted by q1 but acting away from charge q2 that is
Negative sign shows the opposite direction.Hence the two forces are equal but opposite in direction.This is similar to Newton’s third law.
φ = E→.A→
= Eacos Θ
Where E is the magnitude of E→.A is the magnitude of A→ and Θ is the angle between the electric intensity(lines of force)and the normal to the surface (because the direction of a vector area is always along the normal to the surface)
As the surface is placed at right angle to the field.The angle between the normal to the surface and the electric intensity will be Zero i.e. Θ =0
φe = Eacoso= EA
Thus maximum flux will pass through the surface placed at right angle to the electric fields.
When the charge q enclosed b a closed surface is positive the following formula for Gauss ‘s law has been derived as,
φ = – Q/ Ԑo
Where φ is the flux through the closed surface and Ԑo is the permitivity of free space.
If charge q is taken as negative then the formula will become as
φ = – q/ Ԑo
In this case the field lines of forces will enter the closed a surface along the normal’s to the surface while in the case of positive q the field lines were coming out of the closed surface.
Electric P.E. = U = qV.
(ii) Electric potential difference between two points is defined as the work done in moving a unit positive charge from one point to the other keeping the charge in electrostatic equilibrium.
Mathematically it is written as
?V = -Ed or ?V = E.d
where ?V is the potential difference between two points separated by a distance ‘d’ and ‘E’is the electric intensity.
Its Definition:-It is defined as the energy gained or lost by an electron
when it is moved between two points with a potential difference of one volt.
1 eV = 1.6 ×10-19 C×1 volt
=1.6 ×10-19 j
Units:- Its SI unit is volt.
Definition of Volt:- If a work of 1j is done in moving 1C change from one point to the other then the potential difference between two points will be one volt.In other words it can be written as
Negative sigh indicates that charge is being moved from a point of lower potential to a point of higher potential.
Or qE= mg
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