1
$\begingroup$

While reading capacitance I found out this statement that for large C, potential difference is small for a given charge means that a capacitor with large capacitance can hold large amount of charge at relatively small potential difference. "High potential difference implies strong electric field around the conductor". A strong electric field can ionise the surrounding air and discharge...

Please explain the lines in quotes maybe by physical interpretation or mathematically!

$\endgroup$

2 Answers 2

0
$\begingroup$

The key to understanding the quoted segment is understanding what electrical potential really means. The strict mathematical definition of the voltage, $V$, between any two points $A$ and $B$ in space is $V=\oint_SE(r)\cdot ds$, where $S$ is an arbitrary path between $A$ and $B$. As a result, voltage scales linearly with $E(r)$, which means that a higher voltage difference between two points also implies a stronger electric field between them.

If you're not familiar with vector calculus, the intuitive way to think about it is that the voltage difference between two points is proportional to the amount of energy required to move a charged particle from one to the other (for instance, the amount of energy required to "push" a proton from the negatively charged plate to the positively charged plate of a capacitor). It stands to reason that if this voltage is very large, the electric field must be very large as well, because this is what we are "pushing" against. This is all very handwavy, but hopefully it gives you an idea of why large potential differences should also imply the existence of strong electric fields.

$\endgroup$
0
$\begingroup$

Assume that the separation of the plates of the capacitor, $d$, is constant as is the electric field between the plates.

The magnitude of the electric field between the plates, $E$, is equal to the potential gradient, $\frac Vd$ where $V$ is the potential difference between the plates.

Thus as the potential difference is increased between the plates so does the electric field between the plates in proportion.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.