Answer:
Explanation:
Potential energy of the system of charges
= 9 x 10⁹ x [ q₁q₂ / r₁₂ + q₂q₃ / r₂₃ + q₁q₃ / r₁₃ ]
here r₁₂ , r₂₃ , r₁₃ are distance between 1 st and 2 nd charge , 2 nd and 3 rd charge and fist and third charge.
r₁₂ = 8 cm , r₂₃ = 4 cm , r₁₃ = 4 cm.
q₁ = 20 x 10⁻⁹ C , q₂ = - 20 x 10⁻⁹ C , q₃ = 10 x 10⁻⁹ C
Potential energy = 9 x 10⁹ x [ - 400 x 10⁻¹⁸ / .08 + -200x10⁻¹⁸ / .04 + 200 x 10¹⁸ / .04 ]
= 9 x 10⁹ x - 400 x 10⁻¹⁸ / .08
= 45 x 10⁻⁶ J .
b)
Potential at the point of fourth charge due to three charges of 20 nC , - 20 nC and 10 nC at the centre
9 x 10⁹ [ 20 x 10⁻⁹ / .05 + - 20 x 10⁻⁹ / .05 + 10 x 10⁻⁹ / .03 ]
= 9 x 10⁹ x 10 x 10⁻⁹ / .03
= 3000 V .
potential energy of fourth particle = charge x potential
= 3000 x 40 x 10⁻⁹ = 12 x 10⁻⁵ J .
kinetic energy at infinity = 12 x 10⁻⁵ J
1/2 m v² = 12 x 10⁻⁵ J
.5 x 2 x 10⁻¹³ x v² = 12 x 10⁻⁵
v² = 12 x 10⁸
v = 3.46 x 10⁴ m/s
= 9 x 10⁹
