Question

In the figure, OA and OB show the variation of electric potential V at a point due to two point charges $Q_1$ and $Q_2$ with $1/r$ respectively. Here r represents the distance of the point from the two point charges. Identify the nature of the two charges $Q_1$ and $Q_2$.

Hint:

Graphical interpretations are a major recurring staple in CBSE physics; always carefully double-check which specific physical variables are rigidly assigned to the respective coordinate axes.
A line projecting into the first quadrant denotes a positive proportionality constant, while the fourth quadrant strictly denotes a negative one.

Answer 1

Concept:
• The scalar electric potential $V$ uniquely created by an isolated point charge $Q$ at a radial distance $r$ is explicitly given by the formula $V = \frac{1}{4\pi\epsilon_0}\frac{Q}{r}$.

• Consequently, if a mathematical graph is meticulously plotted between the potential $V$ on the y-axis and the inverse distance $1/r$ on the x-axis, the resulting curve is structurally a straight line passing perfectly through the origin.

• The mathematical slope of this specific straight line is strictly proportional to both the magnitude and the fundamental sign of the source charge $Q$.

Step 1: Analyze the mathematical relationship
The governing electrostatic equation is:
\[ V = \left(\frac{Q}{4\pi\epsilon_0}\right) \cdot \frac{1}{r} \]
By comparing this directly to the standard linear equation form $y = mx$, we can easily map the variables:
Here, $y$ strictly corresponds to $V$, and $x$ strictly corresponds to $(1/r)$.
The resulting slope $m$ of the graphed line is therefore definitively equal to the constant term:
\[ m = \frac{Q}{4\pi\epsilon_0} \]
Because the factor $1/(4\pi\epsilon_0)$ is a strictly positive universal constant, the sign of the slope $m$ is completely determined by the physical sign of the charge $Q$.

Step 2: Evaluate Line OA to determine $Q_1$
Looking closely at the graphical lines provided in the figure, we mathematically assess their respective plotted slopes.
The plotted line OA visibly lies entirely within the first quadrant, making an angle of $+60^\circ$ with the positive x-axis.
Because the tangent of an acute positive angle is positive, the mathematical slope of line OA is demonstrably positive.
Since a positive slope implies a strictly positive generating charge according to our derived relation, $Q_1$ must definitively be a positive point charge.

Step 3: Evaluate Line OB to determine $Q_2$
Conversely, the plotted line OB inherently ventures downward into the fourth quadrant, generating negative potential values for positive distances.
It geometrically makes an angle of $-30^\circ$ (or $330^\circ$) with the primary positive x-axis.
The tangent of this specific angle is mathematically negative, firmly meaning the slope of line OB is inherently negative.
Therefore, driven by the same logic, $Q_2$ must definitively be a negative point charge.