Given that c=q/v even if I connect it to a higher V, its charge Q may decrease proportionally right? So why should it damage my capacitor? or will the internal electric field get too high and cause the dielectric to break down? Or would it just get too leaky and then overheat due to greatly increased self heating?

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    \$\begingroup\$ youtube search for exploding capacitors to get a visual representation \$\endgroup\$
    – PlasmaHH
    Jan 21, 2016 at 13:45
  • \$\begingroup\$ You wouldn't want to touch the resulting pieces. The electrolyte is caustic, for starters. \$\endgroup\$
    – bwDraco
    Jan 22, 2016 at 7:27
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    \$\begingroup\$ Only once in my life have I actually blown a cap. It was not pretty, even with the blast contained entirely within the housing of the device (a generic USB hub). That smell will teach you to treat capacitors with respect. \$\endgroup\$
    – bwDraco
    Jan 22, 2016 at 7:41

5 Answers 5


A literal answer is this:


There are three blown capacitors; two can be seen as spirals of grey material still reasonably in situ, the third is nothing more than the base and the internal terminals. They were all rated for 6.3V but, do to a failure in the power regulator, they were connected to a whopping 7.5V. A negligible amount, so one would think, but the outer can of that third capacitor blew off with such force that it punched a hole in a piece of 3mm plastic - about 80mm away - and embedded itself into a battery on the other side.

All that brown stuff is a fibrous material similar to cardboard, and it gets everywhere. I don't know if there's some kind of oil inside the capacitor that dries when it is exposed to air but I do know that it sticks like glue to whatever it lands on.

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    \$\begingroup\$ The capacitors in the foreground have cross-shaped score marks on the top. I would guess that's to help them burst in a controlled manner. That is, they're deliberately designed so as to burst at a high pressure but before they build up to a really colossal pressure. \$\endgroup\$ Jan 21, 2016 at 14:34
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    \$\begingroup\$ @steveverrill They are (or were) all cross-topped versions. The one closest to the front has begun to bulge but had not burst before one of the series inductors burnt out completely, cutting off supply to the caps. It seems that sometimes the damage is done too quickly and the cap blows catastrophically. For example, I've never seen a reverse-polarity capacitor fail with any dignity, it's always been a total eruption. \$\endgroup\$ Jan 21, 2016 at 14:39
  • 1
    \$\begingroup\$ Heres an question with answers confirming it electronics.stackexchange.com/q/7929/50922 . Without the cross markings the capacitor cans would have lasted a while longer but the damage would have been much worse. See how they are designed to fail top downwards, not bottom upwards. I'm a chemical engineer and you can get bursting discs that look just like the tops of capacitors to install on pressure vessels. When you work out how much energy is stored in compressed vapours and divide it by the mass, the ratio is enormous. Hence very high velocities are achieved on bursting. \$\endgroup\$ Jan 21, 2016 at 14:55
  • \$\begingroup\$ I suggest the voltage they were exposed to must have been considerably greater than 7.5V. \$\endgroup\$
    – user207421
    Jan 22, 2016 at 1:16
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    \$\begingroup\$ Man, that was not 'literal'. It was graphic. ;-] \$\endgroup\$ Jan 22, 2016 at 2:26

You have to be careful with these equations.

c=q/v, Q = CV, all look very nice, but they only apply within the limits for which they apply.

For a capacitor, one of the limits is keeping the voltage low enough that the capacitor dielectric stays intact. As you increase the terminal voltage, the electric stress increases across the dielectric, and eventually, it breaks down. When that happens, you don't have a capacitor any more. In the best case you are left with a short circuit or an open circuit. In the worst case you have a lab full of smoke and/or a trip to the ER.

Capacitor manufacturers are quite helpful in printing the maximum voltage their caps will stand before they stop being capacitors. You generally can exceed that a little bit, a few percent, at the cost of capacitor lifetime. If you exceed it by 10s of percent, then you'll find your capacitor lifetime becomes zero.

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    \$\begingroup\$ YES! People don't think about how the components actually work. When you charge the plates of a cap there is a mechanical force exerted in proportion to the electrostatic field strength of the charge. The plates flex. If you overcharge, they bend. When they bend they get closer which only increases the inter-plate attraction and the mechanical force. If they bend enough to touch they dead short which flash boils the electrolyte with the illustrated consequences. \$\endgroup\$
    – Peter Wone
    Jan 22, 2016 at 1:53
  • \$\begingroup\$ @Peter Wone: bending plates of an electrolytic (look at the tags carefully) capacitor? RoTFL \$\endgroup\$ Aug 24, 2016 at 15:22
  • \$\begingroup\$ I know they're rolled up. The mechanical forces still occur and contact is still catastrophic. \$\endgroup\$
    – Peter Wone
    Aug 25, 2016 at 0:08
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    \$\begingroup\$ @Peter Wone: An electromagnetic field, certainly, produces “mechanical forces”, but, given the construction of an electrolytic capacitor, they tend to crush the alumina film (BTW Al₂O₃ is, normally, a strong material) and unlikely bend anything in the process. Of course, (eventual) breakdown of alumina might be assisted by electrostatic attraction, as for any solid insulator, but it has nothing to do with “inter-plate attraction” reasonably defined. \$\endgroup\$ Aug 26, 2016 at 13:48

If you want to know why something is happening in the real world, you need a more complex model than the pure theoretical formula.

How are the capacitors made? They are two thin sheets of electrically conductive material with a thin sheet of electrically insulator material placed between them. The capacitance is given by these sheets' geometry. You need a thinner insulator or a larger surface for higher capacity.

In theory, the insulator does not permit the electrons to flow through it. Materials in real life behave differently. With enough voltage applied, any insulator will be forced to allow electrons to flow through it.

The breakdown voltage where this happens depends on the material, also on its geometry. A thinner sheet of insulator will break down at a lower voltage than a thicker one.

This breakdown phenomenon usually is highly energetic, because the small amount of current will dissipate as heat on the huge resistance of the isolator. This might also be a simplification of the real life phenomenon of overvoltage breakdown. Chemical reactions might also occur which can change the capacitor's behavior.

So, if you want to make a small capacitor of a high capacitance, it will have to be limited to low voltages. High voltage, high capacitance ones are big for this reason.

  • \$\begingroup\$ And for engineers, all this complicated real-world behaviour is simplified by the manufacturers to a voltage rating :D \$\endgroup\$
    – slebetman
    Jan 22, 2016 at 12:24

Per @andy the formula needs to be applied in the correct manner.

per @andy and predicted by @user44635 the capacitor will fail when the voltage is raised beyond some limit.

The way it fails and effects thereof depend on

  • the failure voltage,
  • energy stored (\$\frac{1}{2}CV^2\$ at time of failure),
  • rate of change in charge and voltage,
  • type of capacitor,
  • material and manufacturing defects,
  • environmental factors such as humidity and temperature, attached power sources.

@ceteras adds some useful insights to @user44635 and shows how we must always be aware of both the theory and practical relationships in what we are dealing with.

The effects can be insignificant - a puff of smoke or dangerous, life threatening and catastrophic.

In one incident in the 1960s, a relatively small capacitor - I think it was 33pF or so - (about 150mm by 25mm square) my dad manufactured triggered a lot of collateral damage. A small town of about 100K people was without lights for a week-end. The cap was on either a 33kV or 100kV AC line. It was used as part of a capacitive divider for voltage measurement.

It failed due to design and manufacturing defect. I do not remember if anyone was killed or badly injured. This could easily have been the case.

Per @Loren the calcs work out as follows taking 33kV and 33pF (which is what I seem to remember them being marked as)

\$\frac{1}{2}CV^2 = \frac{1}{2} \times (33 \times 10^{-12}) \times (33 \times 1.4 \times 10^3)^2\$

=~ 35mJ (e&oe thanks @peter @loren )

The factor of 1.4 corrects for RMS->peak voltage, caps tend to fail at the peaks.

Discharge of the cap would take in the region of 1ms yielding 35W (maybe a lot faster).

@ 100kV you get 9 times the energy and power - 320mJ.

The dielectric failed, probably due to an imperfection. The entire town supply (several MVA, even in those days) was redirected towards the cap, air ionised, the rest is history. The hot end would have been a busbar, the ground end was attached to another cap as a divider parallel to a neon panel indicator.

Enough to wake the operator but little else. The contribution of the power line through ionised air, would have lasted a bit longer and done the damage.

In the presence of

  high power
  high voltage 
  high current 
  high energy electrical systems of all forms 

a lot of energy may be stored and released quickly at voltages and currents abnormal for the circuitry.

@Charlie shows a nice low voltage example.

Electrolytic caps are interesting in failure mode since the fluids (often in gels) may boil and cause an explosive failure from the volume of hot gasses now occupying their interior. They may reach temperatures above 100celcius before they explode and release superheated steam.

Engineers need always to be concerned with safety of themselves and others.

Charging a capacitor always has some risk as it may fail even when operated within its rated limits due to manufacturing, handling, environmental or for any other reason.

  • \$\begingroup\$ I'm not getting a lot of power in that capacitor even if the line was 100kv. Now, if it responded by shorting and 100kv tried to flow through it that's quite another matter. \$\endgroup\$ Jan 22, 2016 at 5:03
  • \$\begingroup\$ Hi Loren as we see from @Charlie surprising things happen with caps all the time, and its often (usually?) the attached environment that causes the damage. I'll edit my answer with a bit more detail shortly. \$\endgroup\$
    – ChrisR
    Jan 22, 2016 at 5:37
  • \$\begingroup\$ picofarads are 10^-12, not 10^-6 (micro). Your energy numbers are off by a factor of a million, which is probably why they disagree with @Loren's. Almost certainly anything dramatic that happened was due to the failure resulting in a short, or maybe initiating ionization of the air allowing current to flow from the mains. \$\endgroup\$ Jan 22, 2016 at 9:54

Q=CV so, if capacitance remains constant and you raise the voltage, the charge must increase. Connecting a capacitor to a voltage that exceeds its ratings is asking for a puff of smoke or maybe even some fireworks.

  • \$\begingroup\$ @ChrisR who are you directing the comment to? The dust in your eye comment may need some form of explanation as to its relevance. \$\endgroup\$
    – Andy aka
    Jan 21, 2016 at 11:05
  • \$\begingroup\$ Apologies @andy, I will try to be more careful in future. \$\endgroup\$
    – ChrisR
    Jan 21, 2016 at 11:38
  • \$\begingroup\$ @ChrisR You can write new comments and delete the old comments - it's not a problem and I would recommend you doing this if you think they are misleading. I asked because I wasn't sure what you were driving at. \$\endgroup\$
    – Andy aka
    Jan 21, 2016 at 11:41
  • \$\begingroup\$ I have posted the comment in the form of an answer. \$\endgroup\$
    – ChrisR
    Jan 21, 2016 at 13:00

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