Question

In: Physics

A point charge q2 = -3.9 μC is fixed at the origin of a co-ordinate system...

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.

Solutions

Expert Solution

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 RP

RP = ( d12 + a2 )1/2

= ( 0.0822 + 0.022 )1/2

= 0.0844 m

distance R q3 is RR

Rp = ( d22 + a2 )1/2

= ( 0.0332 + 0.022 )1/2

= 0.03858 m

change in potential energy is dPE = PER - PEP

= 2 k q q1 ( 1/RR - 1/RP )

= 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

----------------------------------------------------------------------------------------------------------------

final potential energy of the system is

PER = k ( q1 q3 / RR + ( q1 q4 / RR ) + 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 ]

PER = 2.726442 J


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