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

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) $800\,\mu\text{C}$.