Advertisements
Advertisements
प्रश्न
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?
Advertisements
उत्तर
Given:
Charge of the ball, q = 4.9 × 10−5 C
Electrical field intensity, E = 2 × 104 N/C
Mass of the ball, m = 100 gm
Force of gravity, Fg = mg
Electrical force, Fe = Eq
The particle moves due to the resultant force of Fg and Fe.
\[R^2 = {F_g}^2 + {F_e}^2 \]
\[ = (0 . 1 \times 9 . 8 )^2 + (4 . 9 \times {10}^{- 5} \times 2 \times {10}^4 )^2 \]
\[ = 0 . 9604 + 96 . 04 \times {10}^{- 2} \]
\[ = 1 . 9208 N\]
\[ \Rightarrow R = 1 . 3859 N\]
Fg = Fe
⇒ tanθ = 1
⇒ θ = 45°
θ is the angle made by the horizontal with the resultant.
Hence, the path of the ball is straight and is along the resultant force at an angle of 45° with the horizontal
Vertical displacement in t = 2 s,
\[y = \frac{1}{2}g t^2 \]
\[ \Rightarrow y = \frac{1}{2} \times 9 . 8 \times 2 \times 2 = 19 . 6 \] m
Both the forces are same.
So, vertical displacement in 2 s = Horizontal displacement in 2 s
Net displacement
\[= \sqrt{\left( 19 . 6 \right)^2 + \left( 19 . 6 \right)^2} = \sqrt{768 . 932} = 27 . 7 \] m
APPEARS IN
संबंधित प्रश्न
A hollow cylindrical box of length 0.5 m and area of cross-section 25 cm2 is placed in a three dimensional coordinate system as shown in the figure. The electric field in the region is given by `vecE = 20 xhati` where E is NC−1 and x is in metres. Find
(i) Net flux through the cylinder.
(ii) Charge enclosed by the cylinder.
The charge on a proton is +1.6 × 10−19 C and that on an electron is −1.6 × 10−19 C. Does it mean that the electron has 3.2 × 10−19 C less charge than the proton?
When the separation between two charges is increased, the electric potential energy of the charges
Consider the situation in the figure. The work done in taking a point charge from P to Ais WA, from P to B is WB and from P to C is WC.
The electric field and the electric potential at a point are E and V, respectively.
The electric field in a region is directed outward and is proportional to the distance rfrom the origin. Taking the electric potential at the origin to be zero,
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?
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).
An electric field \[\vec{E} = ( \vec{i} 20 + \vec{j} 30) {NC}^{- 1}\] exists in space. If the potential at the origin is taken to be zero, find the potential at (2 m, 2 m).
Find the magnitude of the electric field at the point P in the configuration shown in the figure for d >> a.
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).
Which of the following methods can be used to charge a metal sphere positively without touching it? Select the most appropriate.
The surface charge density of a thin charged disc of radius R is σ. The value of the electric field at the center of the disc is `sigma/(2∈_0)`. With respect to the field at the center, the electric field along the axis at a distance R from the center of the disc ______.
In general, metallic ropes are suspended on the carriers taking inflammable materials. The reason is ______.
The electric field intensity produced by the radiations coming from 100 W bulb at 3 m distance is E. The electric field intensity produced by the radiations coming from 50 W bulb at the same distance is:
Five charges, q each are placed at the corners of a regular pentagon of side ‘a’ (Figure).
(a) (i) What will be the electric field at O, the centre of the pentagon?
(ii) What will be the electric field at O if the charge from one of the corners (say A) is removed?
(iii) What will be the electric field at O if the charge q at A is replaced by –q?
(b) How would your answer to (a) be affected if pentagon is replaced by n-sided regular polygon with charge q at each of its corners?
