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Concepts Of Physics MCQ Edition [Volume 2]
Exam
Concepts Of Physics MCQ Edition [Volume 2]
Subject
Physics
Chapter
Electric Field and Potential

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Electric Field and Potential

A ball having a mass of 100 g and a charge of 4.9 × 10^ -5 C is...

A ball having a mass of 100 g100 \text{ g} and a charge of 4.9×10−5 C4.9 \times 10^{-5} \text{ C} is released from rest into an area containing a horizontal electric field of 2.0×104 N C−12.0 \times 10^4 \text{ N C}^{-1}. Where will the ball be located after 2 s2 \text{ s}?

Options

  1. A28 m28 \text{ m} from the starting point along its line of motion
  2. B14 m14 \text{ m} from the starting point along its line of motion
  3. C20 m20 \text{ m} from the starting point along its line of motion
  4. D56 m56 \text{ m} from the starting point along its line of motion

Correct answer

A. 28 m28 \text{ m} from the starting point along its line of motion

Step-by-step solution

Horizontal force acting on the ball:

Fx=qE=(4.9×10−5)×(2.0×104)=0.98 NF_x = qE = (4.9 \times 10^{-5}) \times (2.0 \times 10^4) = 0.98 \text{ N}

Vertical force acting on the ball (taking g=10 m s−2g = 10 \text{ m s}^{-2}):

Fy=mg=0.1×10=1 NF_y = mg = 0.1 \times 10 = 1 \text{ N}

Net force acting on the ball:

F=Fx2+Fy2=(0.98)2+(1)2=0.9604+1=1.9604≈1.4 NF = \sqrt{F_x^2 + F_y^2} = \sqrt{(0.98)^2 + (1)^2} = \sqrt{0.9604 + 1} = \sqrt{1.9604} \approx 1.4 \text{ N}

Acceleration of the ball:

a=Fm=1.40.1=14 m s−2a = \dfrac{F}{m} = \dfrac{1.4}{0.1} = 14 \text{ m s}^{-2}

Distance traveled in t=2 st = 2 \text{ s}:

s=12at2=12×14×(2)2=28 ms = \dfrac{1}{2}at^2 = \dfrac{1}{2} \times 14 \times (2)^2 = 28 \text{ m}
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