A point charge of +5.0 \times 10^{-8}\text{ C} is situated in a vacuum. Taking Coulomb's c… — JAMB UTME

A point charge of $+5.0 \times 10^{-8}\text{ C}$ is situated in a vacuum. Taking Coulomb's constant $k = 9.0 \times 10^9\text{ N}\cdot\text{m}^2\text{C}^{-2}$ , what is the magnitude of the electric field intensity, in $\text{N C}^{-1}$ , at a distance of $0.3\text{ m}$ from the charge?

Answer: 5000 N C^-1

Answer

The magnitude of the electric field intensity at a distance of

0.3\text{ m}

5000\text{ N C}^{-1}

The electric field intensity

E

at a distance

r

from a point charge

q

in free space is given by

E = \frac{k |q|}{r^2}

. Substituting

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

q = 5.0 \times 10^{-8}\text{ C}

, and

r = 0.3\text{ m}

into the expression yields

E = \frac{9.0 \times 10^9 \times 5.0 \times 10^{-8}}{(0.3)^2} = \frac{450}{0.09} = 5000\text{ N C}^{-1}

Step-by-Step Solution

Identify the given values and formula

q = 5.0 \times 10^{-8}\text{ C}

r = 0.3\text{ m}

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

, and

E = \frac{k |q|}{r^2}

Electric field intensity surrounding a point charge depends on the charge magnitude and inversely on the square of the distance.

Calculate the square of the distance

r^2 = (0.3)^2 = 0.09\text{ m}^2

The inverse-square law requires using

r^2

in the denominator.

Substitute values and solve for field intensity

E = \frac{9.0 \times 10^9 \times 5.0 \times 10^{-8}}{0.09} = \frac{450}{0.09} = 5000\text{ N C}^{-1}

Multiplying the terms in the numerator yields

450\text{ N}\cdot\text{m}^2\text{C}^{-1}

, and dividing by

0.09\text{ m}^2

gives

5000\text{ N C}^{-1}

Key Concept

Electric field intensity due to a isolated point charge

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