Question

Two identical charges, 5 muC each are fixed at a distance 8 cm and a charged particle of mass 9 xx 10^(-6) kg and charge -10 muC is placed at a distance 5 cm from each of them and is released. Find the speed of the particle when it is nearest to the two charges.

Answer 1

Answer:

Answer

As there is no external force on the system, so the net change is Mechanical energy is 0.

△U

system

=△K

charge

From the diagram,

U

i

=

5X10

−2

KX(5X10

−6

)(−10X10

−6

)

X2 =−20KX10

−10

J

U

f

=

4X10

−2

KX(5X10

−6

)(−10X10

−6

)

X2 =−25KX10

−10

J

△U

sys

=U

i

−U

f

=5KX10

−10

=4.5J

Now,

△U=△K=1/2mv

2

⟹4.5=0.5X9X10

−6

v

2

v=1000m/s