What is the magnitude of the electric field intensity, in \text{N C}^{-1}, at a point 2.0\… — JAMB UTME

What is the magnitude of the electric field intensity, in $\text{N C}^{-1}$ , at a point $2.0\text{ m}$ away from a isolated point charge of $+4.0 \times 10^{-6}\text{ C}$ in a vacuum? (Take Coulomb's constant $k = 9.0 \times 10^9\text{ N m}^2\text{ C}^{-2}$ )

Answer: 9000 N C^-1

Answer

The magnitude of the electric field intensity is

9000\text{ N C}^{-1}

The electric field intensity

E

produced by a point charge

q

at distance

r

is given by

E = \frac{kq}{r^2}

. Substituting

k = 9.0 \times 10^9\text{ N m}^2\text{ C}^{-2}

q = 4.0 \times 10^{-6}\text{ C}

, and

r = 2.0\text{ m}

yields

E = \frac{9.0 \times 10^9 \times 4.0 \times 10^{-6}}{4.0} = 9000\text{ N C}^{-1}

Step-by-Step Solution

Identify the given physical quantities and formula

q = 4.0 \times 10^{-6}\text{ C}

r = 2.0\text{ m}

k = 9.0 \times 10^9\text{ N m}^2\text{ C}^{-2}

. Formula:

E = \frac{kq}{r^2}

The magnitude of electric field intensity due to a single point charge is given by Coulomb's field law.

Substitute the values and calculate the electric field strength

E = \frac{9.0 \times 10^9 \times 4.0 \times 10^{-6}}{2.0^2} = \frac{36000}{4} = 9000\text{ N C}^{-1}

Perform basic arithmetic simplification to determine the numerical result.

Key Concept

Electric Field Intensity due to a Point Charge

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