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In my book, it's given that:

”Principle of Capacitor: In the capacitor arrangement, the increase in capacitance of a conductor is due to the decrease in potential V (charge Q remains constant) when another conductor is brought near to it. Suppose a metallic plate A fitted to an insulated stand is given the charge +Q so that its potential increases to V. Its capacitance is then

$C=Q/V$

When the plate is connected to the disc of a gold leaf electroscope, the leaves of the electroscope diverge. The divergence of the leaves measures the potential of A. When another metallic plate B fitted to an insulated stand is brought close to A, the divergence of the leaves decreases showing that the potential of A has been lowered.”

My question is why does the potential of a plate decrease when anothe plate is brought near it ? In fact , E(of single plate) =$$ \sigma/2\epsilon_o$$ and E(between 2 equal and oppositely charged plates) = $$\sigma/\epsilon_o$$ Thus we see that electric field between 2 plates is actually more .

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You should remeber that the magnitude of the field is not related to the magnitude of the potential but to the gradient of the potential. The two plates have potenatials with opposite signs so there is a partial cancelation which results in smaller values of potential. Regrading the field, between the plates the potential changes from a positive value to a negative value over a relatively small distance. This means large gradient and so, large field.

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  • $\begingroup$ Okay , but even if the potential turns negative ( as per kq1q2/r) then how does that increase capacitance ? (C=Q/V) Because the absolute value of V increases , the ratio infact decreases. $\endgroup$
    – Debodit Ray
    Commented Oct 29, 2021 at 15:39
  • $\begingroup$ No, the absolute value of the potential decreases. When you add a positive and a negative number you get a value lower than either absolute values of the two numbers. This is the whole point of capacitors as a pair of objects charged with opposite charges. You decrease the potential for a given charge. $\endgroup$
    – nasu
    Commented Oct 29, 2021 at 16:23
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Let's suppose that plate B is grounded. When you bring it close to plate A, current will flow briefly through the grounding wire, and B will be charged by induction.

At that point, the charge on plate A no longer will be uniformly distributed. The mobile charges on plate A and everything connected to it (i.e., charge in the wire and charge in the electroscope) will be attracted toward the opposite charge on plate B and vice versa. Less charge will be available to be seen by the electroscope.

I don't know for sure what will happen if plate B is not grounded. I'm guessing it might be similar, but that the effect would be smaller.

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  • $\begingroup$ Yeah , but the plate is positively charged . So plate B becomes negative by induction , which will repel electrons on plate A , thereby pushing them to the electroscope and increasing the divergence of leaves . $\endgroup$
    – Debodit Ray
    Commented Oct 29, 2021 at 16:15
  • $\begingroup$ @DeboditRay, "pushing [electrons] toward the electroscope" can decrease the divergence of the leaves if the instrument was positively charged to begin with. $\endgroup$
    – Solomon Slow
    Commented Oct 29, 2021 at 16:41

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