Calculate the value of electric field intensity due to a charge of 4μC if the force due to the charge is 8N

1. Analyze the question

We are asked to find the electric field intensity ($E$) given the force ($F$) exerted on a charge and the magnitude of the charge ($q$).

• Force ($F$) = 8 N
• Charge ($q$) = 4 μC (microcoulombs)

2. Explain the concept

The electric field intensity ($E$) at any point in an electric field is defined as the force per unit charge placed at that point. The formula is:

Important Note: Before calculating, you must convert the charge from microcoulombs (μC) to the standard SI unit, Coulombs (C).

• $1 \mu C = 10^{-6} C$
• Therefore, $4 \mu C = 4 \times 10^{-6} C$

3. Show the solution

• Step 1: List the given values. • $F = 8 \, \text{N}$ • $q = 4 \times 10^{-6} \, \text{C}$

• Step 2: Substitute the values into the formula. $$E = \frac{8}{4 \times 10^{-6}}$$

• Step 3: Perform the division. • Divide the numbers: $8 \div 4 = 2$ • Handle the power of ten: $\frac{1}{10^{-6}} = 10^6$

• Step 4: Combine the results. $$E = 2 \times 10^6 \, \text{NC}^{-1}$$

4. Identify the answer

Comparing our result to the given options, the correct value is $2.0 \times 10^6 \, \text{NC}^{-1}$. This matches option A.

5. Tip

Always watch out for prefixes like micro ($\mu = 10^{-6}$), milli ($m = 10^{-3}$), or nano ($n = 10^{-9}$) in JAMB Physics questions. Forgetting to convert these to standard units is the most common reason students lose marks in electricity calculations!