Physics, asked by Saby123, 11 months ago

Hello there Brainliacs, try and solve the following question and earn 50 points.

Two particles A and B. each carrying a charge Q, are held fixed with a separation d between them. A particle C having mass m and charge q is kept at the middle point of the line AB.

(a) If it is displaced through a distance x perpendicular to AB, what would be thebelectric force experienced by it.

(b) Assuming x <<d,
show that this force is proportional to x.

(c) Under whatconditions will the particle C execute simple harmonic motion if it is released after such a small displacement ?

(d)Find the time period of the oscillations if these
conditions are satisfied.

PLZ No Slamming.

Answers

Answered by AwesomeSoul47
5

Answer:

Hey mate here is your answer....

The charge q is displaced by a distance x on the perpendicular bisector of AB.

The charge q is displaced by a distance x on the perpendicular bisector of AB.As shown in the figure, the horizontal component of the force is balanced.

sinθ=xd22+x2

Total vertical component of the force, F'=2Fsinθ

F'=2×14πε0×qQd22+x2×xd22+x2⇒F'=12πε0×qQxd22+x23/2

F'=2×14πε0×qQd22+x2×xd22+x2⇒F'=12πε0×qQxd22+x23/2This is the net electric force experienced by the charge q✅✅.

(b) When x < < d:

F'=12πε0qQxd23 ∵x2<<d22⇒F'=4πε0qQxd3

(c) For the particle to execute simple harmonic motion:

F' = mw2x

F' = mw2x⇒4πε0qQxd3=m2πT2x⇒T2=mπ3ε0d3Qq⇒ T =mπ3ε0d3Qq1/2➡️⬅️⬅️⬅️

thanks

hope it's helpful for you...

Answered by temporarygirl
1

Hola mate

Here is your answer -

The charge q is displaced by a distance x on the perpendicular bisector of AB.  

The charge q is displaced by a distance x on the perpendicular bisector of AB.As shown in the figure, the horizontal component of the force is balanced.

sinθ=xd22+x2

Total vertical component of the force, F'=2Fsinθ

F'=2×14πε0×qQd22+x2×xd22+x2⇒F'=12πε0×qQxd22+x23/2

F'=2×14πε0×qQd22+x2×xd22+x2⇒F'=12πε0×qQxd22+x23/2This is the net electric force experienced by the charge q.

(b) When x < < d:

F'=12πε0qQxd23 ∵x2<<d22⇒F'=4πε0qQxd3

(c) For the particle to execute simple harmonic motion:

F' = mw2x

F' = mw2x⇒4πε0qQxd3=m2πT2x⇒T2=mπ3ε0d3Qq⇒ T =mπ3ε0d3Qq1/2

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