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: The electrostatic potential \(V\) at a distance \(r\) from an isolated point charge \(Q\) in free space is given by Coulomb's law as: \[ V = \frac{1}{4\pi\varepsilon_0} \frac{Q}{r} \] When plotting \(V\) against \(\frac{1}{r}\), the relation is linear and takes the form \(y = mx\), where \(y = V\), \(x = \frac{1}{r}\), and the slope of the line is given explicitly by: \[ m = \frac{1}{4\pi\varepsilon_0} Q \] From this linear dependence, we can determine both the sign (nature) and magnitude of the charges based on the orientation and steepness of the lines.

Step 1: Determining the nature of the charges \(Q_1\) and \(Q_2\).
Analyze the signs of the slopes for lines OA and OB.
- For line OA, the potential \(V\) is positive for positive values of \(\frac{1}{r}\). This means the slope of line OA is positive (\(m_{OA} > 0\)). Since \(\frac{1}{4\pi\varepsilon_0}\) is a positive constant, the charge \(Q_1\) must be positive (\(Q_1 > 0\)).
- For line OB, the potential \(V\) transitions into negative values, meaning the graph lies below the horizontal axis, giving it a negative slope (\(m_{OB} < 0\)). Consequently, the charge \(Q_2\) must be negative (\(Q_2 < 0\)).

Step 2: Finding the ratio of the charges \(\left(\frac{Q_1}{Q_2}\right)\).
Relate the algebraic slopes to the mathematical magnitudes of the charges.
Let the line OA correspond to charge \(Q_1\). Its slope is: \[ \text{Slope of OA} = \frac{1}{4\pi\varepsilon_0} Q_1 \] Let the line OB correspond to charge \(Q_2\). Taking its absolute magnitude, its slope is: \[ \left|\text{Slope of OB}\right| = \frac{1}{4\pi\varepsilon_0} |Q_2| \] Dividing the equation for the slope of OA by the equation for the magnitude of the slope of OB yields: \[ \frac{\text{Slope of OA}}{\left|\text{Slope of OB}\right|} = \frac{\frac{1}{4\pi\varepsilon_0} Q_1}{\frac{1}{4\pi\varepsilon_0} |Q_2|} = \left|\frac{Q_1}{Q_2}\right| \] Hence, the value of the magnitude of their ratio is directly proportional to the ratio of their respective slopes.