Question

Plot a graph showing the variation of coulomb force (F) versus (1/r²), where r is the distance between the two charges of each pair of charges: (1μC, 2μC) and (2μC – 3μC). Interpret the graphs obtained.

Answer 1

According to Coulomb's law:

\(F = K\frac{q_1q_2}{r^2}\)

\(= K\frac{1\mu C \times 2\mu C}{r^2}\)

⇒ \(F = 2K \left(\frac 1{r^2}\right)\)

To draw graph of F versus 1/r²

On comparing above equation with y = mx where m is the slope

We get, Slope of equation m_A ​= 2K  ....(1)

Both charges will repel each other, as they are of same sign, hence slope is positive.

\(F' = K\frac{q_1'q_2'}{r^2}\)

\(= K\frac{1\mu C \times (-3\mu C)}{r^2}\)

⇒ \(F = -3K \left(\frac 1{r^2}\right)\)

To draw graph of F′ versus 1/r²

On comparing above equation with y = mx, where m is the slope

We get, Slope of equation m_B​ = −3K    ....(2)

Both charges are of opposite sign, therefore they will attract each other, that's why slope is negative. 

y = mx is the equation of straight line

From equation (1) and (2), we observe that

Magnitude of slope of B is greater than the magnitude of slope of A 

i.e. mB​ > mA​

Hence, graph between F and 1/r²​ will be straight line where line B has higher slope than line A as shown in figure above.

Answer 2

F = 1/(4πε₀) (q₁q₂)/r²

The graph between F and 1/r² is a straight line of slope 1/(4πε₀) q₁ q₂ passing through origin.

Since, magnitude of the slope is more for attraction, therefore, attractive force is greater than repulsive force.