Question

A 2 – gram object, located in a region of uniform electric field \(\vec{E}=(300\ N\ C^{-1})\hat{i}\) carries a charge Q. The object released from rest at x = 0 , has a kinetic energy of 0.12 J at x = 0.5 m. Then Q is

• A. 400 μc
• B. -400 μc
• C. 800 μc
• D. -800 μc

Answer 1

The Correct Option is C

Step 1: Recall the work-energy theorem

The work done on an object is equal to the change in its kinetic energy. In this case, the work done on the object by the electric field is converted into its kinetic energy. The work done by the electric field $ \mathbf{E} $ is given by:

$W = Q \cdot E \cdot d$

where:
- $ Q $ is the charge of the object,
- $ E $ is the electric field strength,
- $ d $ is the displacement of the object in the direction of the electric field.

The kinetic energy gained by the object is given by:

$K.E. = \frac{1}{2} m v^2$

where $ m $ is the mass of the object, and $ v $ is its velocity at a distance $ x = 0.5 $ m.

Step 2: Use the work-energy theorem

We are told that the object starts from rest, so its initial kinetic energy is zero. The kinetic energy at $ x = 0.5 $ m is 0.12 J. Therefore, the work done on the object is equal to this change in kinetic energy:

$W = K.E. = 0.12 \, \text{J}$

Thus, the work done by the electric field is:

$W = Q \cdot E \cdot x$

Substitute the known values:
- $ E = 300 \, \text{N/C} $,
- $ x = 0.5 \, \text{m} $,
- $ W = 0.12 \, \text{J} $.

We can now solve for $ Q $:

$0.12 = Q \cdot 300 \cdot 0.5$

$0.12 = 150 Q$

$Q = \frac{0.12}{150} = 0.0008 \, \text{C} = 800 \, \mu\text{C}$

Final Answer: The charge $ Q $ is $ 800 \, \mu\text{C} $, which matches option (C).

Answer 2

The work done by the electric field on the charged object is equal to the change in kinetic energy. Since the object starts from rest, its initial kinetic energy is zero. The final kinetic energy is given as $0.12\,\text{J}$.

Work done $W = \Delta KE = KE_f - KE_i = 0.12\,\text{J} - 0\,\text{J} = 0.12\,\text{J}$.

The work done is also given by $W = Fd$, where $F$ is the force on the object and $d$ is the distance it moves in the direction of the force. The force on a charged object in an electric field is given by $F = QE$. In this case, the force and displacement are both in the x-direction, so $d = 0.5\,\text{m}$.

$W = QEd$

$0.12\,\text{J} = Q(300\,\text{N/C})(0.5\,\text{m})$

$Q = \frac{0.12\,\text{J}}{(300\,\text{N/C})(0.5\,\text{m})} = \frac{0.12}{150} \text{C} = 0.0008\,\text{C} = 800\,\mu\text{C}$

The correct answer is (C) $800\,\mu\text{C}$.