Question

Two charges of equal magnitudes and at a distance
rexert a force F on each other. If the charges are
halved and distance between them is doubled them
the new force on each charge is :
(a) F/8
(6) F/4
(c) 4F
(d) F/16​

Answer 1

Answer:

• Option D is the Correct Answer.

Explanation:

We have been given that two charges of equal magnitudes and at a distance

r exert a force F on each other. If the

charges are halved and distance between them is doubled them.

• Since we know that original force applied on any object with two different charges is given by F = k(q1*q2/r²)
• Original Force: F = k(q1*q2/r²) Where k is constant & q1 and q2 are charges.

Now, Let's follow the question's Statement:

→ Let Required force be F' .

• New Force : F' = k [q/2*q/2 ÷ (2r²)²]
• F' = k [ q²/4 ÷ 4r⁴) ]
• F' = k [q/2*q/2 ÷ (2r²)²]
• F' = k [ q²/(4r⁴ * 4) ]

After Comparison, we have:

→ F' = 1/16 F

• Therefore, the required force is F/16 and hence, Option (D) is the correct answer.

Answer 2

Option D is correct answer.

Explanation:

It is given that two charges of equal magnitudes and at a distance

(r) exert a force (F) on each other. If the

charges are halved and distance between them is doubled them.

Then,

• Original Force: F = k(q× q/r²)
• Original Force: F = k(q× q/r²)

⭐ Let new force be F' .

Now, According to the Question!

• F' = k {q/2×q/2 / (2r²)²}
• F' = k { q²/4 /4r⁴) }
• F' = k {q²/(4r⁴ × 4) }

After Solving the Equation, we have

• F' = 1/16 F