Question

In the figure, OA and OB show the variation of electric potential V at a point due to two point charges Q₁ and Q₂   with `1/r` respectively. Here r represents the distance of the point from the two point charges.

1. Identify the nature of the two charges Q₁ and Q₂.
2. What is the value of `((Q_1)/(Q_2))?` Justify your answer.

Answer 1

Core Principle:

The electric potential V at a distance r from a point charge Q is given by: 

V = `(kQ)/r`

= `(kQ) . 1/r`

This equation is in the form y = mx, where y = V and x = `1/r`. 

Therefore, the slope (m) of the line represents kQ.

(I) Nature of the Charges Q₁ and Q₂:

Charge Q₁, (Line OA): This line lies in the positive V region (above the horizontal axis). Since the potential is positive, Q₁ is a positive charge.

Charge Q₂ (Line OB): This line lies in the negative V region (below the horizontal axis). Since the potential is negative, Q₂ is a negative charge.

(II) The slope of a line on a graph is equal to the tangent of the angle (tan θ) it makes with the positive horizontal axis.

Slope₁ = kQ₁

= tan (60°)

= `sqrt3`

Since line OB is below the axis, it represents a negative slope:

Slope₂ = kQ₂

= tan (−30°)

= `-1/sqrt3`

The ratio of the charges:

`((Q_1)/(Q_2)) = (kQ_1)/(kQ_2)`

= `(tan 60°)/(-tan 30°)`

= `(sqrt3)/(-1/sqrt3)`

= `-(sqrt3 xx sqrt3)`

= −3