Question

A point charge q₂ = -3.9 μC is fixed at the origin of a co-ordinate system as shown. Another point charge q₁ = 4.5 μC is is initially located at point P, a distance d₁ = 8.2 cm from the origin along the x-axis

Point R located at d₂ = 3.3 cm

The charge q₂ is now replaced by two charges q₃ and q₄ which each have a magnitude of -1.95 μC, half of that of q₂. The charges are located a distance a = 2 cm from the origin

What is the potential energy of the system composed of the three charges q₁, q₃, and q₄, when q₁ is at point R? Define the potential energy to be zero at infinity.

Answer 1

given

q₂ = -3.9 μC

= - 3.9 x 10^-6 C

q₁ = 4.5 μC

= 4.5 x 10^-6 C

distance d₁ = 8.2 cm

= 0.082 m

Point R located at d₂ = 3.3 cm

= 0.033 m

q₃ and q₄ which each have a magnitude of -1.95 μC,

= -1.95 x 10^-6 C

half of that of q₂

a = 2 cm

= 0.02 m

R at point P is R_P

R_P = ( d₁² + a² )^1/2

= ( 0.082² + 0.02² )^1/2

= 0.0844 m

distance R q₃ is R_R

R_p = ( d₂² + a² )^1/2

= ( 0.033² + 0.02² )^1/2

= 0.03858 m

change in potential energy is dPE = PE_R - PE_P

= 2 k q q₁ ( 1/R_R - 1/R_P )

= 2 x 9 x 10⁹ x (-1.95 x 10^-6 ) x 4.5 x 10^-6 x ( 1/0.0844² - 1/0.3858² )

dPE = 83.9459 J

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final potential energy of the system is

PE_R = k ( q₁ q₃ / R_R + ( q₁ q₄ / R_R ) + q₅ q₃ / 2 a )

= 9 x 10⁹ x [( 4.5 x 1.95 x 10^-12 / 0.0844 ) + ( 4.5 x 1.95 x 10^-12 / 0.0844 ) + ( 1.95 x 1.95 x 10^-12 / 0.04 ) ]

= 9 x 10⁹ x [ 1.03969 x 10^-10 + 1.03969 x 10^-10 + 9.5 x 10^-11 ]

PE_R = 2.726442 J