Question

Two positive charges of magnitude q are placed at the ends of a side

Answer 1

Answer:

The kinetic energy at the center of square is \dfrac{1}{4\pi\epsilon_{0}a}2q(1-\dfrac{1}{\sqrt{5}})4πϵ0a12q(1−51)

Explanation:

Given that,

Square side = 2 a

Charge = q

According to figure,

The electric potential at P

V_{p}=\dfrac{1}{4\pi \epsilon_{0}}\times q(\dfrac{1}{a}+\dfrac{1}{a}-\dfrac{1}{a\sqrt{5}}-\dfrac{a}{a\sqrt{5}})Vp=4πϵ01×q(a1+a1−a51−a5a)

V_{p}=\dfrac{1}{4\pi\epsilon_{0}a}\times 2q(1-\dfrac{1}{\sqrt{5}})Vp=4πϵ0a1×2q(1−51)

The electric potential at O

V_{o}=\dfrac{1}{4\pi \epsilon_{0}}\times q(\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{2}})Vo=4πϵ01×q(21+21−21−21)

V_{o}=0Vo=0

The difference is

V_{p}-V_{o}=\dfrac{1}{4\pi\epsilon_{0}a}2q(1-\dfrac{1}{\sqrt{5}})Vp−Vo=4πϵ0a12q(1−51)

The total energy of the charge q at point P

E= \dfrac{1}{4\pi\epsilon_{0}a}2q(1-\dfrac{1}{\sqrt{5}})E=4πϵ0a12q(1−51)

Here, The kinetic energy is zero.

The total energy of the charge q at point O

E= \dfrac{1}{4\pi\epsilon_{0}a}2q(1-\dfrac{1}{\sqrt{5}})E=4πϵ0a12q(1−51)

Here, The potential energy is zero

Hence, The kinetic energy at the center of square is \dfrac{1}{4\pi\epsilon_{0}a}2q(1-\dfrac{1}{\sqrt{5}})4πϵ0a12q(1−51)