Question

Plot a graph showing the variation of coulomb force $\left(F\right)$ versus $\frac{1}{r^2}$ where $r$ is the distance between the two charges of each pair of charge $(1\mu C,2\mu C)$ and $(1\mu C,-3\mu C)$. Interpret to that graphs obtained.

Answer 1

$\text{Step 1: Calculation of force for first pair of charges}$ $1\mu C$ and $2\mu C$

According to Coulomb's law:

$F=K\frac{q_1{q}_2}{r^2}$ $=K\frac{1\mu C\times 2\mu C}{r^2}$

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

To draw graph of $F$ versus $1/r^2$

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

We get, Slope of equation $m_A=2K$ $....\left(1\right)$

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

$\text{Step 2: Calculation of force for second pair}$ $2\mu C$ and $-3\mu C$

$F^{\prime }=K\frac{q_1^{\prime }{q}_2^{\prime }}{r^2}$ $=K\frac{1\mu C\times -3\mu C}{r^2}$

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

To draw graph of $F^{\prime }$ versus $1/r^2$

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

We get, Slope of equation $m_B=-3K$ $....\left(2\right)$

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

$\text{Step 3: Comparing slopes and drawing the graph}$

$y=mx$ is the equation of straight line

From equation $\left(1\right)$ and $\left(2\right)$, we observe that

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

i.e. $m_B>m_A$

Hence, graph between F and$\frac{1}{r^2}$ will be straight line where line $B$ has higher slope than line $A$ as shown in figure above.