Question

Two identical +ve, charges are at the ends of straight line AB and another identical +ve charge is placed at C such that AB = BC, A, B and C being on the same line. Now the force on A​

Answer 1

Answer:

increased by kq²/4r²

Explanation:

  (A)<-------r------->(B)<-------r-------->(C)

   q                       q                       q

   

In between two point charges,

               F = k(q₁q₂)/r²  

In the given cases everything remains same except the distance between the charges.

 Case1:   force is in-between charge at A and charge at B.

         Force = kqq/AB²  = kq²/r²

 

  Case2: force is in-between charge at A and charge at C   &  charge at A and charge at B.

Force(net) = F₁₃ + F₁₂ = kq₁q₃/(2r)²  +  kq₁q₂/r²

           = kq²/(2r)² + kq²/r²

Change in force = final - initial

           = [ kq²/(2r)² + kq²/r² ] - [ kq²/r² ]

           = kq²/(2r)²

Hence the force is increased by kq²/4r².

 

Statement 1:  Force is increased by kq²/4r².

Statement 2: Net force = kq²/(2r)² + kq²/r²

                                      = (5/4) kq²/r²

Statement 3:  Net force is (5/4)th of the initial force.

Statement 4: Force is increased by 25%.

Answer 2

Answer:

Given :-

• Two identical positive charges are at the ends of straight line AB and another identical positive charges are placed at C such that AB = AC ,A,B, C being on the same line .Now the force on A

To find :-

• Force on A will be

1. Increased
2. decreased
3. remains same.
4. not predicted.

Solution :-

• Here when C is initially not present the force on A will be

$= \frac{1}{4\pi \: e0} \times \frac{q1 \times q2}{ {r}^{2} }$

Consider ,

• Case i) when C is added the net force of A becomes

$= (\frac{1}{4\pi \: e0} \times \frac{q1 \times q2}{ {r}^{2} } + \frac{1}{4\pi \: e0} \times \frac{q1 \times q2}{2 {r}^{2} }$

• Since,

• Here
• q1=q2=q.

Hence,

• $= \frac{ {q}^{2} }{4\pi \: e0} )( \frac{1}{r {}^{2} } + \frac{1}{4 {r}^{2} }$

Then,

• F= 5q^2/4×16 pi e0r^2

Therefore ,

• Here F is increased by 1/4pie0×q^2/4r^2.

Option A is the correct answer for your question.

Hope it helps u mate .

Thank you .