Question

A soap bubble is charged to a potential of $16V$, if its radius is doubled, then new potential of the bubble will be

Answer 1

The radius of soap bubble increases because of outward energy following up at the bubble because of charging

Formula used:
$V = \dfrac{q}{c}$
$V =$ Potential
$q =$ Charge
$c =$ Capacitance

Complete step by step answer:
Given, ${V_1} = 16V$
Radius $= r$
Now we have to use the formula nd we can write it as,

Also, it is given that Radius $= 2r$ and Potential $= {V_2}$
Again we use the formula and we can write it as

Now we have to divide (1) with (2)
$\dfrac{{16}}{{{V_2}}} = \dfrac{{\dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{q}{r}}}{{\dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{q}{{2r}}}}$
$\dfrac{{16}}{{{V_2}}} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{q}{r} \times \dfrac{{4\pi {\varepsilon _0}}}{1}\dfrac{{2r}}{q}$
Cancel out the same term and we get,
$\dfrac{{16}}{{{V_2}}} = \dfrac{{2r}}{r}$
On dividing we get
$\dfrac{{16}}{{{V_2}}} = 2$
On rewriting we get,
${V_2} = \dfrac{{16}}{2}$
the radius is doubled then new potential of the bubble is ${V_2} = 8V$

Additional information:
• The presence of impurities at the floor of, or dissolved in, a substance without delay impacts the floor tension of the liquid.
• The floor tension of water, for instance, will increase whilst pretty soluble impurities are delivered. Yes, adding salt to water does increase the floor tension of water, although no longer through any considerable quantity.
• However, experiments finished with a saltwater display that surface tension virtually increases while salt is introduced to herbal water.
• If the floor is agitated to break up the surface tension, then the needle will rapid sink.

Note: Bubble will amplify due to the fact that the charged debris always disseminated on it reasons them to repel every different due to the electrostatic energy. This will appear to each nice and negatively charged bubble due to the price on it.
In stability, the strength introduced because of floor strain is equal to the amount of powers due to excess air strain inner the bubble and the electrical pressure due to charge,
Hence, whilst the cleaner bubble is given a bad charge, then its radius will increment.