# Capacitors with same capacitance but different voltage ratings

Suppose we have two capacitors that have same capacitance (same dielectric material) but different voltage ratings. Let both capacitors each be fully charged to their maximum voltages. From formula $$Q=CV$$ (fixing $$C$$ as constant), capacitor 1 has charge $$Q_1$$ and voltage $$V_1$$; capacitor 2 has charge $$Q_2$$ and voltage $$V_2$$. What makes this change in voltage without change in capacitance value? Since from $$V= E d$$, if we say electric field $$E$$ or separation between plates $$d$$ as the reason for the voltage variation, then it means that $$C$$ also changes, because of the relationship $$C= \varepsilon A/d$$, i.e. $$E$$ depends on $$\varepsilon$$, the dielectric constant.

For the same dielectric material, is it possible to have same capacitance and different voltage rating?

• "What makes this change in voltage without change in capacitance ?" You connect the capacitor to a power supply set to give a higher voltage between its terminals? Commented May 16 at 8:04
• (a) "Yes, after charging the two capacitor[s]" But you do the charging by connecting the capacitors across power supplies! (b) "Two cap. has two different voltages across it according to its rating" Your grammar is confused: are you talking about two capacitors or one capacitor? A capacitor's rating is the maximum voltage that you may safely apply to it. (c) I'm afraid that I don't understand what your difficulty is. Commented May 16 at 8:33
• "What makes this change in voltage without change in capacitance?" If the capacitance is the same, then it's different charge. Not sure what you're asking here. Commented May 16 at 8:45
• You've charged the capacitors to different voltages: call them $V_A$ and $V_B$. Therefore the electric fields strengths have magnitudes $E_A=\tfrac {V_A}d$ and $E_B=\tfrac {V_B}d$ and the charges are $Q_A=CV_A$ and $Q_B=CV_B$. The capacitors are identical, so $C$ is the same for each. $C$ is unaffected by $E$, because for a typical dielectric, $\epsilon$ is unaffected by $E$. Commented May 16 at 11:54
• "For the same dielectric material, is it possible to have same capacitance and different voltage rating?". Yes ; making $d$ larger will enable the capacitor to stand a larger voltage without the dielectric 'breaking down' (conducting when it should be insulating). But to keep the capacitance the same we'd have to make $A$ larger (e.g. if we doubled $d$ we'd have to double $A$). Commented May 16 at 12:06

If you connect the two capacitors in parallel then they have the same voltage across them but they carry different charges - the charge on each capacitor is in proportion to its capacitance, so that $$\frac {Q_1}{C_1} = \frac {Q_2}{C_2}$$.
If you connect the two capacitors in series then they carry the same charge but they have different voltages - the voltage across each capacitor is in inverse proportion to its capacitance, so that $$V_1 C_1 = V_2 C_2$$.