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The radius of a soap bubble whose potential $16\,V$ is doubled. The new potential of the bubble will be:
(A) $2\,V$
(B) $4\,V$
(C) $8\,V$
(D) $16\,V$

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Last updated date: 18th Jul 2024
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Answer
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Hint The new potential of the bubble can be determined by using the electric potential produced by the point charge $Q$, by using this formula and the condition which is given in the question, the new potential developed in the bubble can be determined.

Useful formula
The electric potential produced by the point charge can be determined by,
$V = \dfrac{{kQ}}{R}$
Where, $V$ is the potential developed in the soap bubble, $k$ is the constant, $Q$ is the point charge which is placed in the centre of the bubble and $R$ is the radius of the soap bubble.

Complete step by step solution
Given that,
The initial potential of the soap bubble is, $V = 16\,V$,
Now,
The electric potential produced by the point charge can be determined by,
$V = \dfrac{{kQ}}{R}\,................\left( 1 \right)$
By substituting the potential developed in the soap bubble in the above equation (1), then the above equation (1) is written as,
$16 = \dfrac{{kQ}}{R}\,...................\left( 2 \right)$
Now, the radius of the soap bubble is doubled, then the equation (1) is written as,
$V = \dfrac{{kQ}}{{2R}}$
By rearranging the terms in the above equation, then the above equation is written as,
$V = \dfrac{1}{2}\left( {\dfrac{{kQ}}{R}} \right)$
By substituting the equation (2) in the above equation, then the above equation is written as,
$V = \dfrac{1}{2}\left( {16} \right)$
By dividing the terms in the above equation, then the above equation is written as,
$V = 8\,V$
Thus, the above equation shows the new potential developed in the soap bubbles when the radius is doubled.

Hence, the option (C) is the correct answer.

Note The potential developed is directly proportional to the charge which is placed in the centre of the circle and the potential developed is inversely proportional to the radius of the circle. As the charge increases, the potential developed also increases.