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
1) 100 mC
2) 200 mC
3) 300 mC
4) 35√10 C
[Explain properly with a solution]
Plzz answer it a
and fast
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Let charge on each coin is qGiven, 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
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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