Question

Charges of $2\mu C$ and $-3\mu C$ are placed at two points A and B separated by distance of 1 m. The distance of the point from A where net potential is zero is ______.

• A. 0.667 m
• B. 0.5 m
• C. 0.4 m
• D. 0.6 m

Hint:

The internal zero-potential point always lies closer to the charge with the SMALLER magnitude. $2\mu\text{C}$ is smaller than $3\mu\text{C}$, so $x$ must be less than $0.5$ m. Option (c) is the only logical choice!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We have two point charges of opposite signs separated by 1 meter. We must find the specific location between them where their individual electric potentials perfectly cancel each other out to zero.

Step 2: Detailed Explanation:
Electric potential ($V$) is a scalar quantity. The net potential at any point is the simple algebraic sum of the potentials from each individual charge.
$V_{\text{net}} = V_A + V_B = 0$
The formula for electric potential from a point charge is $V = \frac{kQ}{r}$.
Let the point of zero potential be located at a distance $x$ from charge A.
Therefore, its distance from charge B will be $(1 - x)$.
We are given:
$Q_A = +2 \mu\text{C}$
$Q_B = -3 \mu\text{C}$
Set up the equation for net zero potential:
$\frac{k Q_A}{x} + \frac{k Q_B}{(1 - x)} = 0$
Substitute the charge values (ignoring the $\mu\text{C}$ unit prefix and the constant $k$ as they will perfectly cancel out on both sides):
$\frac{2}{x} + \frac{-3}{1 - x} = 0$
Move the negative term to the right side of the equation:
$\frac{2}{x} = \frac{3}{1 - x}$
Cross-multiply to solve for $x$:
$2(1 - x) = 3x$
$2 - 2x = 3x$
Bring all $x$ terms to one side:
$2 = 3x + 2x$
$5x = 2$
$x = \frac{2}{5} \text{ m}$
$x = 0.4 \text{ m}$

Step 3: Final Answer:
The distance is 0.4 m, matching option (c).