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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?
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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
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