Question

Q. Two Identical coins having similar charges are placed 4.5 m apart on a table. Force of repulsion between them is 49/9 N. The value of charge on each coin is :-

1) 100 mC

2) 200 mC

3) 300 mC

4) 35√10 C

Answer 1

Let charge on each coin is q

Given, Force = 49/9 N
Speration between charges , r = 4.5m

∵ Force = KQq/r² [ from Coulombs law]
∴ 49/9 = 9 × 10⁹ × q × q/(4.5)²
⇒ 49/9 = 9 × 10⁹ × q²/(4.5)²
⇒ 49 × (4.5)² × 10⁻⁹/9 × 9 = q²
⇒(7)² × (4.5)² × 10 × 10⁻¹⁰/(9)² = q²
Taking square root both sides,
⇒ 7 × 4.5 × √10/9 × 10⁻⁵ = q
⇒ 3.5√10 × 10⁻⁵ = q
⇒ 35√10 × 10⁻⁶ = q

Hence, charge on each coin is 35√10 μC
It seems you did mistake in typing.
Answer should be 35√10 μC , Correct Option is (4) 35√10 μC

Answer 2

Hello Dear.

Here is the answer---


There is a little Mistakes in the Question.
In the Question, Option 4 has the units μC instead of C.

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Answer--- 4 Option,  35√10 μC.

Explanation---

Given Conditions,

          Distance between the two coins(r) = 4.5 m
 
  Force of Repulsion between the Charges(F) = 49/9 N.


Using the Coulomb Law,

  F = kq²/r²

Where,
      
     k = Constant
        = 9 × 10⁹

Putting the Values in the Formula,

 49/9 = 9 × 10⁹ × q/(4.5)²

⇒  q = 36√10 × 10⁻⁶ C

We know, 10⁻⁶ = μ

    Thus, q = 36√10 μC

∴ Thus, the Charge on each coin is 36√10 μC.

Hence Option 4 is correct.


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Hope it helps.

Have a nice day.