Advertisements
Advertisements
प्रश्न
A particle of mass 1 g and charge 2.5 × 10−4 C is released from rest in an electric field of 1.2 × 10 4 N C−1. Find the electric force and the force of gravity acting on this particle. Can one of these forces be neglected in comparison with the other for approximate analysis?
Advertisements
उत्तर
Given:
Charge of the particle, q = 2.5 × 10−4 C
Initial velocity, u = 0
Electric field intensity, E = 1.2 × 104 N/C
Mass of the particle, m = 1 g = 10−3 kg
Distance travelled, s = 40 cm = 4 × 10−1 m
Electric force,
\[F_e = qE\]
\[ \Rightarrow F_e = 2 . 5 \times {10}^{- 4} \times 1 . 2 \times {10}^4 = 3 \] N
Force of gravity,
\[F_g = \text{ mg }\]
\[ \Rightarrow F_g = 9 . 8 \times {10}^{- 3} \] N
APPEARS IN
संबंधित प्रश्न
The figure shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio?
Consider a system of n charges q1, q2, ... qn with position vectors `vecr_1,vecr_2,vecr_3,...... vecr_n`relative to some origin 'O'. Deduce the expression for the net electric field`vec E` at a point P with position vector `vecr_p,`due to this system of charges.
Why does a phonograph record attract dust particles just after it is cleaned?
In some old texts it is mentioned that 4π lines of force originate from each unit positive charge. Comment on the statement in view of the fact that 4π is not an integer.
When the separation between two charges is increased, the electric potential energy of the charges
The electric field and the electric potential at a point are E and V, respectively.
Electric potential decreases uniformly from 120 V to 80 V, as one moves on the x-axis from x = −1 cm to x = +1 cm. The electric field at the origin
(a) must be equal to 20 Vcm−1
(b) may be equal to 20 Vcm−1
(c) may be greater than 20 Vcm−1
(d) may be less than 20 Vcm−1
Consider a uniformly charged ring of radius R. Find the point on the axis where the electric field is maximum.
A wire is bent in the form of a regular hexagon and a total charge q is distributed uniformly on it. What is the electric field at the centre? You may answer this part without making any numerical calculations.
A particle of mass 1 g and charge 2.5 × 10−4 C is released from rest in an electric field of 1.2 × 10 4 N C−1. How long will it take for the particle to travel a distance of 40 cm?
A ball of mass 100 g and with a charge of 4.9 × 10−5 C is released from rest in a region where a horizontal electric field of 2.0 × 104 N C−1 exists. (a) Find the resultant force acting on the ball. (b) What will be the path of the ball? (c) Where will the ball be at the end of 2 s?
12 J of work has to be done against an existing electric field to take a charge of 0.01 C from A to B. How much is the potential difference VB − VA?
An electric field of 20 NC−1 exists along the x-axis in space. Calculate the potential difference VB − VA where the points A and B are
(a) A = (0, 0); B = (4 m, 2m)
(b) A = (4 m, 2 m); B = (6 m, 5 m)
(c) A = (0, 0); B = (6 m, 5 m)
Do you find any relation between the answers of parts (a), (b) and (c)?
Consider the situation of the previous problem. A charge of −2.0 × 10−4 C is moved from point A to point B. Find the change in electrical potential energy UB − UA for the cases (a), (b) and (c).
Assume that each atom in a copper wire contributes one free electron. Estimate the number of free electrons in a copper wire of mass 6.4 g (take the atomic weight of copper to be 64 g mol−1).
When 1014 electrons are removed from a neutral metal sphere, the charge on the sphere becomes ______.
Two similar spheres having +Q and -Q charges are kept at a certain distance. F force acts between the two. If at the middle of two spheres, another similar sphere having +Q charge is kept, then it experiences a force in magnitude and direction as ______.
The Electric field at a point is ______.
- always continuous.
- continuous if there is no charge at that point.
- discontinuous only if there is a negative charge at that point.
- discontinuous if there is a charge at that point.
