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. What is the value of $Q_1/Q_2$? Justify your answer.

Hint:

A negative angle physically denotes a clockwise angular measurement strictly from the primary positive x-axis.
Always remember basic trigonometric values like $\tan(60^\circ)$ and $\tan(30^\circ)$ to quickly execute slope calculations without relying on external tables.

Answer 1

Concept:
• As established, the slope of the $V$ versus $1/r$ graph serves as a direct mathematical proxy for the magnitude and sign of the source charge.

• The precise slope of any straight line graphed on a Cartesian plane can be easily calculated by taking the trigonometric tangent of the angle it makes with the positive x-axis.

• By mathematically finding the exact ratio of the two calculated slopes, we can systematically discover the exact ratio of the two corresponding point charges.

Step 1: Establish the Slope-Charge Proportionality
From the standard electrostatic potential formula, the linear slope $m$ of the $V$ versus $1/r$ graph is formally given by:
\[ m = \frac{V}{(1/r)} = \frac{Q}{4\pi\epsilon_0} \]
This fundamentally means that the charge $Q$ is strictly and directly proportional to the calculated slope $m$:
\[ Q = m \cdot (4\pi\epsilon_0) \]

Step 2: Calculate the Slope for Charge $Q_1$
For the first point charge $Q_1$, the corresponding line is denoted as OA.
The graph explicitly indicates that line OA makes an angle of $\theta_1 = 60^\circ$ strictly with the positive x-axis.
The mathematical slope $m_1$ is computed using the standard tangent function:
\[ m_1 = \tan(60^\circ) = \sqrt{3} \]

Step 3: Calculate the Slope for Charge $Q_2$
For the second point charge $Q_2$, the corresponding line is denoted as OB.
The graph visually shows that line OB makes an angle of $30^\circ$ below the positive x-axis.
In standard mathematical convention, an angle measured clockwise from the positive x-axis is rigorously taken as negative, so $\theta_2 = -30^\circ$.
The mathematical slope $m_2$ is computed similarly:
\[ m_2 = \tan(-30^\circ) = -\tan(30^\circ) = -\frac{1}{\sqrt{3}} \]

Step 4: Determine the Final Charge Ratio
We algebraically divide the derived charge equations to meticulously find the explicit ratio of the point charges.
Because the $4\pi\epsilon_0$ constant perfectly cancels out during division, the charge ratio is strictly identical to the slope ratio:
\[ \frac{Q_1}{Q_2} = \frac{m_1 \cdot 4\pi\epsilon_0}{m_2 \cdot 4\pi\epsilon_0} = \frac{m_1}{m_2} \]
Substitute the exact trigonometrically calculated slope values securely into the fraction:
\[ \frac{Q_1}{Q_2} = \frac{\sqrt{3}}{-1/\sqrt{3}} \]
Simplifying this fractional mathematical expression mechanically yields the final answer:
\[ \frac{Q_1}{Q_2} = \sqrt{3} \times (-\sqrt{3}) = -3 \]
This robustly concludes both the exact magnitude ratio and firmly confirms the opposing signs of the charges.