Calculate E at 0.3m from a 2μC charge. Direction rules.

Electric Field Due to a Point Charge — E = kq/r²

Question

A point charge of $q = 2\,\mu\text{C}$ is placed in free space. Find the electric field intensity at a point $P$ located 0.3 m from the charge. Also state the direction of the field.

Solution — Step by Step

We have $q = 2\,\mu\text{C} = 2 \times 10^{-6}\,\text{C}$ and $r = 0.3\,\text{m}$ .

The value of Coulomb’s constant is $k = 9 \times 10^9\,\text{N\,m}^2\text{C}^{-2}$ .

The electric field due to a point charge is:

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

This formula gives the magnitude of the field. Direction is handled separately — never mix them into the arithmetic.

E = \frac{9 \times 10^9 \times 2 \times 10^{-6}}{(0.3)^2}

E = \frac{9 \times 10^9 \times 2 \times 10^{-6}}{0.09}

E = \frac{18 \times 10^3}{0.09} = 2 \times 10^5\,\text{N/C}

$E = 2 \times 10^5\,\text{N/C}$

Since $q$ is positive, the field at $P$ points radially outward — away from the charge, directly along the line joining $q$ to $P$ .

If $q$ were negative, the direction reverses: the field would point toward the charge. The magnitude formula stays identical either way.

Why This Works

Coulomb’s law tells us that the force between two charges falls off as $1/r^2$ — this is the inverse square law, the same pattern as gravity. The electric field is just the force per unit positive test charge ( $E = F/q_0$ ), so it inherits that same $1/r^2$ dependence.

The $r^2$ in the denominator is critical. Every time you double the distance, the field becomes four times weaker. This is why problems often ask you to compare fields at different distances — the ratio trick is faster than recalculating each time.

We use $k = 9 \times 10^9$ instead of the full form $k = \dfrac{1}{4\pi\varepsilon_0}$ for speed. Both are correct; in board exams, either notation is accepted.

Alternative Method

In JEE, when the charge is given in $\mu\text{C}$ and distance in cm or m, convert units first and keep powers of 10 separate. This prevents arithmetic errors under pressure.

We can also express $k = \dfrac{1}{4\pi\varepsilon_0}$ and use $\varepsilon_0 = 8.85 \times 10^{-12}\,\text{C}^2\text{N}^{-1}\text{m}^{-2}$ :

E = \frac{q}{4\pi\varepsilon_0 r^2} = \frac{2 \times 10^{-6}}{4\pi \times 8.85 \times 10^{-12} \times 0.09}

This route is longer for numericals but necessary when questions give you $\varepsilon_0$ directly or ask about fields inside a medium (where $\varepsilon_0$ gets replaced by $\varepsilon_0\varepsilon_r$ ).

For this problem, the $k = 9 \times 10^9$ shortcut gives the same answer: $2 \times 10^5\,\text{N/C}$ .

Common Mistake

Students often forget to square the distance. Writing $E = kq/r$ instead of $E = kq/r^2$ is the single most common error in NCERT-type numericals. Here, using $r = 0.3$ instead of $r^2 = 0.09$ gives $E = 6 \times 10^4\,\text{N/C}$ — off by a factor of 3.33. Always check: the denominator must have $r^2$ , not $r$ .

A second trap: mixing up the unit of charge. $2\,\mu\text{C}$ means $2 \times 10^{-6}\,\text{C}$ , not $2 \times 10^{-3}\,\text{C}$ (that would be milliCoulombs). In CBSE 2023 and JEE Main 2022, several students lost marks purely on this conversion.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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