Question

The force between two point charges in air is100N.calculate the force if the distance between them is increased by 50percent

Answer 1

Given:

Force between two point charges in air, F= 100N

To Find:

Force between given two point charges if the distance between them is increased by 50%

Solution:

We know that,

• Force F between two charges is given by

$\pink{\boxed{\bf{F=\frac{kq_{1}q_{2}}{r^{2}}}}}$

where

q₁ and q₂ are the magnitude of two charges

r is the distance between two charges

k is dielectric constant of medium

$\rule{190}{1}$

Let the magnitude of charge of given two point charges be q₁ and q₂ respectively and r be the distance between two charges

So, on using above formula, we get

$\longrightarrow\rm{F=\dfrac{kq_{1}q_{2}}{r^{2}}}$

$\longrightarrow\rm{100=\dfrac{kq_{1}q_{2}}{r^{2}}}$

$\longrightarrow\rm{\dfrac{kq_{1}q_{2}}{r^{2}}=100}-------(1)$

$\rule{190}{1}$

Now, on increasing the distance between two charges by 50%, new distance will be

$\longrightarrow\rm{r'=r+50\%\:of\:r}$

$\longrightarrow\rm{r'=r+\dfrac{50}{100}\times r}$

$\longrightarrow\rm{r'=r+\dfrac{1}{2}\times r}$

$\longrightarrow\rm{r'=r+\dfrac{r}{2}}$

$\longrightarrow\rm{r'=\dfrac{2r+r}{2}}$

$\longrightarrow\rm{r'=\dfrac{3r}{2}}$

$\rule{190}{1}$

Let the new force between two charges be F'

So,

$\longrightarrow\rm{F'=\dfrac{kq_{1}q_{2}}{(r')^{2}}}$

$\longrightarrow\rm{F'=\dfrac{kq_{1}q_{2}}{\bigg(\dfrac{3r}{2}\bigg)^{2}}}$

$\longrightarrow\rm{F'=\dfrac{kq_{1}q_{2}}{\dfrac{9r^{2}}{4}}}$

$\longrightarrow\rm{F'=\dfrac{4kq_{1}q_{2}}{9r^{2}}}$

$\longrightarrow\rm{F'=\dfrac{4}{9}\bigg(\dfrac{kq_{1}q_{2}}{r^{2}}\bigg)}$

From (1), we get

$\longrightarrow\rm{F'=\dfrac{4}{9}(100)}$

$\longrightarrow\rm{F'=\dfrac{400}{9}}$

$\longrightarrow\rm\green{F'=44.44\:N}$

Hence, force between given two point charges if the distance between them is increased by 50% is 44.44 N.