Question

Two charges q1 and q2 are placed 30cm apart. A third charge q3 is moved along the arc of a circle of radius 40cm frm C to D. The change in the potential energy of the system is q3 k/4 pi epsilon0, where 'k' is :

a) 8 q1.

b) 6 q1.

c) 8 q2.

d) 6 q2.

Answer 1

Answer: (C). $k = 8 q_{2}$

Explanation:

Given that,

Distance between two charges d = 30 cm

Radius r = 40 cm

The change in potential energy of the system is

$U = \dfrac{q_{3}k}{4\pi\epsilon_{0}}$

We know that ,

The potential energy

$U = \dfrac{1}{4\pi\epsilon_{0}}[\dfrac{q_{1}q_{2}}{d}]$

When the charge $q_{3}$ at point C then,

Using Pythagoras theorem

The distance is 0.5 m between charges $q_{2}$ and  $q_{3}$

The potential energy

$U_{C} = \dfrac{1}{4\pi\epsilon_{0}}[\dfrac{q_{1}q_{3}}{0.4\ m}+\dfrac{q_{2}q_{3}}{0.5\ m}]$

When the charge  $q_{3}$ at point D then,

The distance is 0.1 m between charges $q_{2}$ and  $q_{3}$

The potential energy

$U_{D} = \dfrac{1}{4\pi\epsilon_{0}}[\dfrac{q_{1}q_{3}}{0.4\ m}+\dfrac{q_{2}q_{3}}{0.1\ m}]$

Now, the change in the potential energy

$U = U_{D}- U_{C}$

$\dfrac{q_{3}k}{4\pi\epsilon_{0}} = \dfrac{1}{4\pi\epsilon_{0}}[\dfrac{q_{2}q_{3}}{0.1\ m}- \dfrac{q_{2}q_{3}}{0.5\ m}]$

$\dfrac{q_{3}k}{4\pi\epsilon_{0}}= \dfrac{q_{2}q_{3}}{4\pi\epsilon_{0}}[\dfrac{1}{0.1\ m} - \dfrac{1}{0.5\ m}]$

$k = 8 q_{2}$

Hence, k is $8q_{2}$.