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 \(\frac{1}{r}\) respectively. Here \(r\) represents the distance of the point from the two point charges.

(I) Identify the nature of the two charges \(Q_1\) and \(Q_2\).
(II) What is the value of \(\left(\frac{Q_1}{Q_2}\right)\)? Justify your answer.

Hint:

In graphs of \(V\) vs \(\frac{1}{r}\): - A positive slope represents a positive point charge. - A negative slope represents a negative point charge. - The steepness (absolute value of the slope) is directly proportional to the magnitude of the charge.

Answer 1

Concept: For a point charge, $V = \dfrac{Q}{4\pi\varepsilon_0}\cdot\dfrac{1}{r}$. This has the form $y = mx$ with $y = V$ and $x = \dfrac{1}{r}$, so each line through the origin has slope $m = \dfrac{Q}{4\pi\varepsilon_0}$, and $Q = 4\pi\varepsilon_0\, m$.

Step 1: Reading the angles from the figure.
OA is drawn $60^\circ$ above the $\dfrac{1}{r}$ axis, so its inclination is $\theta_A = +60^\circ$. OB is drawn $30^\circ$ below the axis, so $\theta_B = -30^\circ$.

Step 2: Signs of the charges (part I).
$\tan 60^\circ > 0$ gives a positive slope for OA, hence $Q_1$ is a positive charge. $\tan(-30^\circ) < 0$ gives a negative slope for OB, hence $Q_2$ is a negative charge.

Step 3: Value of $\dfrac{Q_1}{Q_2}$ (part II).
$$\frac{Q_1}{Q_2} = \frac{m_A}{m_B} = \frac{\tan 60^\circ}{\tan(-30^\circ)} = \frac{1.732}{-0.577} = -3$$ Justification: both lines are drawn for the same medium, so the factor $\dfrac{1}{4\pi\varepsilon_0}$ is common and cancels, leaving the ratio of the slopes.

Final Answer: $Q_1$ is positive, $Q_2$ is negative, and $\dfrac{Q_1}{Q_2} = -3$ (the magnitudes are in the ratio $3:1$).