According to relation $V$ equals $\frac{W}{Q}$ , $4V$ is inversely proportional to $Q$ but according to relation $Q$ equals to $CV$ potential is directly proportional to charge, what is the difference between these two cases?
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3 Answers
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These two equations describe completely different things.
$V = W/Q$ says that if you have a test charge $Q$, and you want to move it from place-1 to place-2, and it takes an amount of work $W$ to do it, then the potential (voltage) at place-2 is higher than that at place-1 by an amount $V$. The equation may make it may look like $V$ depends on $Q$, but it does not: if you double the charge, it takes twice as much work to move it, and the $V$ remains the same. The potential is a function of the pre-existing electric field, and the equation simply tells you how much work it takes to move a given amount of charge around in that field.
$Q=CV$ says that the potential difference across a capacitor is proportional to the amount of charge on each plate of the capacitor, and defines the capacitance, $C$ as the constant of proportionality. It is much less general than the first equation: there are many situations where the concept of capacitance is not applicable. Moreover, the charge referred to in the equation is the charge that produces the field, not one that experiences it.
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$\begingroup$ In general ,what would be the relation between charge and potential ;Are they directly proportional to each other? $\endgroup$ Commented Jan 13, 2018 at 4:25
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$\begingroup$ They are not proportional. Potential varies smoothly. Charge can be distributed however you want. For example, the potential due to a point charge $q$ is $V=kq/r$, where $k$ is a constant and $r$ is the distance from the point charge. You can get the potential due to other arrangements of charge by superposition of multiple point charges. $\endgroup$– Ben51Commented Jan 13, 2018 at 4:36
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$\begingroup$ By relation q equals to cv ,if you increase the potential then charge increases ,can it mean that in case of capacitors they are directly proportional? $\endgroup$ Commented Jan 13, 2018 at 4:54
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$\begingroup$ Not everywhere on the capacitor. If you define V to be zero midway between the capacitor plates, then the potential on the plates is proportional to the charge on them. But there is not charge in the space between the plates, and there the potential varies linearly. $\endgroup$– Ben51Commented Jan 13, 2018 at 4:57
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To talk about potential, let's first discuss the electric field. The force on a charge due to an electric field $\vec{E}$ is given by $\vec{F} = q\vec{E}$, where $q$ is the charge of the particle. The way we think about the electric field is that at any point in space, it has a given value. This value is called the potential $V$.
The electric field does work on a particle to push it towards a point that has a lower potential. If two points have a big difference in potential, the field will apply a large force to move the particle, whereas only a small amount of force will be applied for small differences in potential (imagine a ball rolling down a large hill vs. a small hill). Thus, the electric field strength is defined as the rate of change of potential, or mathematically, $\vec{E} = - \nabla V$.
Note that the value of $V$ by itself is arbitrary; only the difference in potential between two points is what's significant.
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What is "the" relationship between potential and charge.
Voltage is sometimes called potential
Charge refers to a quantity that seems to be constant. Ie the charge of an electron.
Q above called charge refers to how many electrons
Delta q = C delta V For a capacitor the noted constant farads
"The relationship between potential and charge" ... in a capacitor ... The higher the value in Farads the lesser potential each element of charge contains in the capacitor.
So some capacitors are better at stacking electrons say one at a time so the capacitor contains a larger voltage releationship than other capacitors.
You have to remember that the capacitor needs to be integrated over time to relate to the work.