A charge of $4.6 × 10^{-5}$C is placed in an electric field of intensity $3.2 × 10^4$ $Vm^{-1}$. What is the force acting on the electron?

Analysis

To solve this problem, we need to understand the relationship between an electric field, the charge placed within it, and the resulting force exerted on that charge.

Concept

The electric field intensity ($E$) is defined as the force ($F$) per unit charge ($q$) acting on a charge placed in an electric field. The formula is:

To find the force ($F$), we rearrange the formula:

Where:

• $F$ = Force (measured in Newtons, N)
• $q$ = Charge (measured in Coulombs, C)
• $E$ = Electric field intensity (measured in Volts per meter, $Vm^{-1}$ or Newtons per Coulomb, $NC^{-1}$)

Step-by-Step Solution

1. Identify the given values: • Charge ($q$) = $4.6 \times 10^{-5} \, \text{C}$ • Electric field intensity ($E$) = $3.2 \times 10^4 \, \text{Vm}^{-1}$

2. Apply the formula: $$F = (4.6 \times 10^{-5}) \times (3.2 \times 10^4)$$

3. Perform the calculation: • Multiply the coefficients: $4.6 \times 3.2 = 14.72$ • Multiply the powers of 10: $10^{-5} \times 10^4 = 10^{-1}$ • Combine them: $14.72 \times 10^{-1} = 1.472 \, \text{N}$

4. Rounding: Rounding $1.472 \, \text{N}$ to one decimal place gives us approximately 1.5 N.

Conclusion

Comparing our result to the given options, the value matches option A.

Exam Tip: When dealing with scientific notation in JAMB Physics, always multiply the numbers first, then handle the powers of 10 separately. This prevents errors with decimal placement. Also, remember that $1.472$ is very close to $1.5$, so don't be discouraged if your exact calculation isn't listed; look for the closest approximation!