© All Rights reserved @ LearnWithDash
Step 1: Understand the Given Data
• Mass of the bullet, $m = 2\,\text{g} = 2 \times 10^{-3}\,\text{kg}$
• Charge on the bullet, $q = 2\,\mu \text{C} = 2 \times 10^{-6}\,\text{C}$
• Final speed of the bullet, $v = 10\,\text{m/s}$
• Initial speed, $u = 0\,\text{m/s}$ (starting from rest)
Step 2: Recall the Relation Between Kinetic Energy and Potential Difference
When a charged particle is accelerated from rest through a potential difference $V$, the work done on the particle by the electric field (which is $qV$) is converted into its kinetic energy (which is $\tfrac{1}{2} m v^2$). Hence,
$ \tfrac{1}{2} m v^2 = qV. $
Step 3: Substitute the Values
Rearranging to find $V$, we get
$ V = \dfrac{\tfrac{1}{2} m v^2}{q}. $
Substitute the numerical values:
$ V = \dfrac{\tfrac{1}{2} \times (2 \times 10^{-3}\,\text{kg}) \times (10\,\text{m/s})^2}{2 \times 10^{-6}\,\text{C}}. $
Step 4: Perform the Calculations
• First, calculate the numerator (kinetic energy):
$ \tfrac{1}{2} \times 2 \times 10^{-3}\,\text{kg} \times (10\,\text{m/s})^2 = 1 \times 10^{-3} \times 100 = 0.1\,\text{J}. $
• Now, divide by the charge:
$ V = \dfrac{0.1}{2 \times 10^{-6}} = \dfrac{0.1}{2\times 10^{-6}} = 0.1 \div (2 \times 10^{-6}) = 50 \times 10^{3} = 50\,\text{kV}. $
Step 5: State the Final Answer
Hence, the bullet must be accelerated through a potential difference of 50 kV to acquire a speed of 10 m/s.
