Let's solve the problem step-by-step: **Given:** - Acceleration of the electron, \( a = 2 \times 10^{10} \) m/s\(^2\) - Charge of electron, \( e = 1.6 \times 10^{-19} \) C - Mass of electron, \( m = 9.1 \times 10^{-31} \) kg We need to find the magnitude of the electric field (\( E \)). --- ### **Step 1: Write the equation of motion for the electron** Newton's second law for the electron: \[ F = ma \] The force on the electron due to electric field: \[ F = eE \] --- ### **Step 2: Equate the two forces** \[ eE = ma \] \[ E = \frac{ma}{e} \] --- ### **Step 3: Substitute the values** \[ E = \frac{(9.1 \times 10^{-31} \text{ kg})(2 \times 10^{10} \text{ m/s}^2)}{1.6 \times 10^{-19} \text{ C}} \] --- ### **Step 4: Calculate the numerator** \[ 9.1 \times 10^{-31} \times 2 \times 10^{10} = 18.2 \times 10^{-21} \] --- ### **Step 5: Divide by the denominator** \[ E = \frac{18.2 \times 10^{-21}}{1.6 \times 10^{-19}} \] \[ E = \frac{18.2}{1.6} \times 10^{-21+19} \] \[ E = 11.375 \times 10^{-2} \] \[ E = 0.11375 \text{ kN/C} \quad \left(\text{since } 1 \text{ kN/C} = 1000 \text{ N/C}\right) \] But let's check the options: - A) 3.11 kN/C - B) 5.21 kN/C - C) 7.12 kN/C - D) 1.13 kN/C Let's express the answer in kN/C: \[ E = 1.1375 \times 10^2 \text{ N/C} = 0.11375 \text{ kN/C} \] Wait, let's check the calculation again: \[ E = \frac{(9.1 \times 10^{-31}) \times (2 \times 10^{10})}{1.6 \times 10^{-19}} \] Calculate numerator: \[ 9.1 \times 2 = 18.2 \] \[ 10^{-31} \times 10^{10} = 10^{-21} \] So numerator: \( 18.2 \times 10^{-21} \) Denominator: \( 1.6 \times 10^{-19} \) So, \[ E = \frac{18.2 \times 10^{-21}}{1.6 \times 10^{-19}} \] \[ = \frac{18.2}{1.6} \times 10^{-21 + 19} \] \[ = 11.375 \times 10^{-2} \] \[ = 0.11375 \text{ N/C} \] Wait, this does not match the options, which are in kN/C (should be much larger). Let's try again carefully: \[ E = \frac{ma}{e} \] \[ = \frac{9.1 \times 10^{-31} \times 2 \times 10^{10}}{1.6 \times 10^{-19}} \] \[ = \frac{18.2 \times 10^{-21}}{1.6 \times 10^{-19}} \] \[ = \frac{18.2}{1.6} \times 10^{-21+19} \] \[ = 11.375 \times 10^{-2} \] But \( 10^{-21+19} = 10^{-2} \), so: \[ E = 11.375 \times 10^{-2} \text{ N/C} = 0.11375 \text{ N/C} \] But the options are in kN/C (1000 N/C). So, if we multiply by 1000 to get kN/C: \[ 0.11375 \text{ N/C} = 0.00011375 \text{ kN/C} \] This is still not matching the options. Let's check the calculation for any mistakes. #### Let's try with units: \[ m = 9.1 \times 10^{-31} \text{ kg} \] \[ a = 2 \times 10^{10} \text{ m/s}^2 \] \[ e = 1.6 \times 10^{-19} \text{ C} \] Multiply numerator: \[ 9.1 \times 2 = 18.2 \] \[ 10^{-31} \times 10^{10} = 10^{-21} \] So numerator: \( 18.2 \times 10^{-21} \) Divide by denominator: \[ 1.6 \times 10^{-19} \] So, \[ E = \frac{18.2 \times 10^{-21}}{1.6 \times 10^{-19}} \] \[ = \frac{18.2}{1.6} \times 10^{-21+19} \] \[ = 11.375 \times 10^{-2} \] \[ = 0.11375 \text{ N/C} \] This seems correct, but the numbers are too small. Let's check the value of electron mass: The mass of electron is actually \( 9.1 \times 10^{-31} \) **kg**, and the charge is \( 1.6 \times 10^{-19} \) **C**. But let's try calculating with more precision: \[ E = \frac{9.1 \times 10^{-31} \text{ kg} \times 2 \times 10^{10} \text{ m/s}^2}{1.6 \times 10^{-19} \text{ C}} \] \[ = \frac{18.2 \times 10^{-21}}{1.6 \times 10^{-19}} \] \[ = \frac{18.2}{1.6} \times 10^{-2} \] \[ = 11.375 \times 10^{-2} \] \[ = 0.11375 \text{ N/C} \] There's a mistake. Let's check the calculation for the exponent: \[ 10^{-31} \times 10^{10} = 10^{-31 + 10} = 10^{-21} \] \[ 18.2 \times 10^{-21} \div 1.6 \times 10^{-19} = 18.2/1.6 \times 10^{-21 - (-19)} = 11.375 \times 10^{-2} \] The exponent arithmetic is correct. But the options are in kN/C (1000 N/C), and our answer is 0.11375 N/C, which is 0.00011375 kN/C. **It seems there is a mismatch in the numbers or possibly a typo in the problem or options.** --- ### **Final Answer:** **Calculation gives:** \[ \boxed{0.11375 \text{ N/C}} \] **But none of the options match this result. Please check if the acceleration provided in the question is correct, or if there's a typo in the options.** --- **If you have any corrections or further details, please let me know!**