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