A spark is produced between two insulated surfaces maintained at a potential difference of 5×106V. If the energy output is 10−5J, the charge transferred during the spark is :
A
5×1011C
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B
5×10−11C
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C
2×1012C
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D
2×10−12C
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Solution
The correct option is D2×10−12C We know that if a charge is transferred by potential diffrence of ΔV, Then, Work done W=qΔV ⇒10−5=q×5×106 ⇒q=15×10−11 ⇒q=2×10−12C