Two 0.5 kg charged objects are held in place, side by side, 0.2 meters apart, on

Question

Two 0.5 kg charged objects are held in place, side by side, 0.2 meters apart, on a flat, horizontal, frictionless surface. Both objects have the same charge on them due to 3.125x10^13 excess electrons on each of their surface.

a) what is the charge on each of the charged object (in C)?

b) what is the force on the right object from the left object (in N)?

c) what is the final acceleration of the right object if it is released and allowed to move freely (in m/s^2)?

d) in which direction will the right object move (to the left or right)?

Please explain your answers!

Explanation / Answer

Firstly we will find the magnitude of charge in coloumb.

Charge on 1 electron=1.6*10—¹9 C

Charge due to 3.125*10¹³ electrons=

3.125*10¹³ * 1.6*10—¹9=5*10^-6 coloumb or 5 micro coloumb. This is the amount of charge in coloumb on each object.

B) :Force on right due to charge of left=F;Kq1q2/r²

Where k=constant =9*10^9

Q1 qnd q2 are charges and r is distance between them

F=9*10^9 * 5*10^-6 * 5*10^-6/(0.2)²

Force =225*10—³/4*10—²

Force=225*10—¹ /4

Force =22.5/4=5.625

C) :if the object on right is released then acceleration

=force /mass

Acceleration =5.625/0.5=

11.25m/s²

D):since the nature of charges is similar i.e both are negative charges,there will be repulsive force between the two.

Therefore the object in right will move away from the object in left or we say that it will move in right direction.

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