A parallel-plate capacitor is made of two circular plates, each of which has a diameter of 2.50 × 10−3 m. The plates of the capacitor are separated by a space of 1.40 × 10−4 m. If the potential difference of 0.12 V is removed from the circuit and the circuit is allowed to discharge until the charge on the plates has decreased to 70.7 percent of its fully charged value, what will the potential difference across the capacitor be in V?

  The potential difference is

Explanation:

From the question we are told that

   The diameter of the of both plates  is  

    The distance of separation is  

     The potential difference is

    The new charge on the plate is  % of

Where is the original charge on the capacitor

Generally the charge on the capacitor is mathematically represented as

Where C is the capacitance of the capacitor which is mathematically represented as

Where   is the permittivity of free space which is a constant with value  

A is the area which is mathematically represented as

substituting values

Now

So

Generally

Here

Making V the subject in the above formula

Substituting values

fgravity=10x10=100n case 1

fgravity=30x20=600n case 2

6 times more

hello, i am papaguy.

i have found an answer to your question but use it as information.

information: because weak acids/bases have low concentration of hydrogen ions, the addition of water has a large impact on the ph. 2. a difference in 1 unit is a 10x difference in concentration for example; a liquid with ph of 3 is 10x more acidic than a liquid with a ph of 4. therefore, a liquid with a ph of 3 is stronger.

best of luck!