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

[Explain properly with a solution]

Plzz answer it a
and fast

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.
The units of Option 4 in the question will be μC instead of C.

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Answer--   4) Option - 35√10 μC

Explanations --

  Given Conditions---

 Distance between the two charges(r) = 4.5 m.
 Force of Repulsion(F0 = 49/9 N.

Using the Coulomb Law,

    Force of Repulsion between the Charges = kq²/r²
 Where,
    k = 8.99 × 10⁹
       = 9  × 10⁹   [Approx]

q = charge

Now, Putting the Given Values in the Formulas,

      49/9 = 9  × 10⁹ × q²/(4.5)²
  ⇒   9 × 10⁹ × q² = 12.25

  On Solving this,
         q = 35√10 × 10⁻⁶ C

We know, 10⁻⁶ = μ

 Thus, q = 35√10 μC.

∴The value of Charge in each coin is 35√10 μC

Hence, Option 4)  is correct.


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Hope it helps.

Have a Marvelous Day.