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Two positive charges of magnitude q are ...

Two positive charges of magnitude q are placed at the ends of a side (side 1 ) of square of side 2a. Two negative cahrges of the same magnitude ar kept at the other comers. Starting from rest, if the charge Q moves from the middle of side (1) to the centre of square , its kinetc energy at the centre of square is

A

`1/(4 pi in_(0)) (2qQ)/a (1 - 1/(sqrt(5))) `

B

zero

C

`1/(4 pi in_(0)) (2qQ)/a ( 1+ 1/(sqrt(5))`

D

`1/(4 pi in_(0)) (2qQ)/a ( 1- 2/sqrt(5))`

Text Solution

Verified by Experts

The correct Answer is:
A
Refer figure,
AC = AD = a
` AE = AF = sqrt(a^(2)+ (2a)^(2))`
`= sqrt(5) a `
`BC = BD= BE = BF = sqrt2 a`
Potential at point A is
` V_(A) ( 1/(4 pi in_(0)) q/(AC) + 1/(4 pi in_(0)) q/(AD) -1/(4 pi in_(0)) q/(AE) - 1/(4 pi in_(0)) q/(AF))`
`= 1/(4 pi in_(0) q/a + 1/( 4 pi in_(0)) q/a - 1/(4 pi in_(0)) q/(sqrt(5a)) - 1/(4 pi in_(0)) q/(sqrt(5a)))`
` V_(A) = 2 (1/(4 pi in_(0) q/a)) -2 1/(4 pi in_(0)) q/(sqrt(5a)) = 1/(4 pi in_(0)) (2q)/a ( 1- 1/sqrt(5))`
Potenital at point B is
` V_(B) ( 1/(4 pi in_(0)) q/(BD) + 1/(4 pi in_(0)) q/(BE) -1/(4 pi in_(0)) q/(AE) - 1/(4 pi in_(0)) q/(BF))`
` (1/(4 pi in_(0)) q/sqrt(2a) + 1/(4 pi in_(0)) q/(sqrt2a)- 1/(4 pi in_(0)) q/(sqrt(2 a)) - 1/(4 pi in_(0)) q/(sqrt(2a))`
`V_(B)= 2 1/( 4 pi in_(0) q/(sqrt(2)a)) - 2 (1/(4 pi in_(0)) q/(sqrt(2)a) = 0`
` therefore V_(A)-V_(B) = 1/(4 pi in_(0) (2q)/a (1-1/sqrt5)`
Kinetic energy at the centre of square of charge Q
= Work done on charge Q from A to B is
` W = Q (V_(A)-B-V_(B)) = 1/( 4pi in_(0)) ( 2 qQ)/a ( 1- 1/sqrt(5))`
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Knowledge Check

  • Two positive charges of magnitude q are placed at the ends of a side ( side 1) of a square of side 2a . Two negative charges of the same magnitude are kept at the other corners . Staring from rest , n a charge Q moves from the middle of side 1 to the centre of square , its kinetic energy at the centre of square is -.

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  • The electric field at the centre of a square having equal charge q at each of the coner (side of square a )

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