A capacitor is the combination of two metal plates (conductors) having opposite charges separated by distance "d" apart. Don't assume that the metal plates should be rectangular plates with some infinite distance.
Our usual capacitor is just so small enough within the size of a finger, where the conducting material rolled down in a cylindrical shape with some non conducting or dielectric material in between them.
Let us discuss with mathematics,
The two plates have opposite charges, let say, "+Q" on the first plate and "-Q" on the second plate. So, the potential difference between these plates can be calculated as the work done required to bring a unit positive charge from -Q to +Q [from the definition of potential]
$$ \int_{V_-}^{V_+}\,dV = V = V_+ - V_- = - \int_{-}^{+} \vec{E}\cdot \vec{dl} \,\,\,...eq.(1)$$
We don't assume anything about the Shapes of the plates, so Electric field is just given by the definition as, $$ \vec{E} = \frac{1}{4\pi\epsilon_0} \int \rho \frac{\hat{r}}{r^2} d\tau $$ where $\rho$ is the volume charge density and integral is over the volume "V".
Electric field is proportional to both Potential "V" and the charge on the each capacitor "Q" and the ratio is some constant which is defined as the capacitance. $$ C = \frac{Q}{V} $$
Let us try to calculate the capacitance of some simple known shapes, where assumptions are easy to make.
Capacitance of Parallel Plate Capacitor
First one is the parallel plate capacitor, where Electric field is directed from positive charge to negative charge plate and assumed to be uniform in the direction. So, it can be taken out from the integral in equation (1) and integral sign is canceled by the negative sign and the equation becomes, $$ V = E \int_0^d \,dl = Ed $$ From our definition of Capacitance, $$ C = \frac{Q}{V} = {Q}{E d} $$ And, the Electric field in the region between parallel plate capacitors is given by, $$ E = \frac{\sigma}{\epsilon_0} $$ where $\sigma = \frac{Q}{A}$ is the surface charge density.
Then, $$ C = \frac{Q}{\frac{Qd}{A\epsilon_0}} = \frac{A\epsilon_0}{d} $$ which is determined only by the sizes, shapes, and separation distance of the two conductors.
Capacitance of Cylindrical capacitor
Second is the cylindrical capacitors, where inside is solid cylinder with positive charge with radius "a" and outside is hollow cylinder with bigger radius "b" (b>a). Let us take a point "r" in the "in between region a<r<b "
The Electric field in this region is given by making use of Gauss law in cylindrical symmetry as, $$ \vec{E} = \frac{\lambda}{2\pi r \epsilon_0} \hat{r}$$
Then potential difference is calculated by, $$ - V = - \int_a^b \vec{E} \cdot \vec{dr} $$ where $\vec{dr} $ is directed from "a" to "b" i.e. radially outwards. So, it becomes $$ V = \frac{\lambda}{2\pi \epsilon_0} \int_a^b \,\frac{1}{r} dr = \frac{\lambda }{2\pi\epsilon_0}\ln{(\frac{b}{a})}$$
where again the integral sign canceled by the negative sign(slightly different from the previous - Here we started with finding $-V = V_ - - V_+$ because the limits will be easy and all are measured from the centre - otherwise $\vec{dl} $ will be directed from -Q to +Q which will have limits 0 to "b-a" where logarithmic function not finite).
Thus we get the capacitance of the cylindrical capacitor as, $$ C = Q/V = \frac{\lambda{l} 2\pi\epsilon_0}{\lambda \ln{(\frac{b}{a})}} = \frac{2\pi\epsilon_0{l}}{\ln{(\frac{b}{a})}} $$
Capacitance of spherical capacitor
Finally, let us consider a spherical capacitor with two concentric spherical metal shells with radii a and b. Inner shell with +Q and outer shell with -Q charge.
In the same way, the Electric field in the region is given by, $$\vec{E} = \frac{Q}{4\pi\epsilon_0 r^2} \hat{r} $$ vector "r" points radially outwards.
Potential difference is calculated by, (integration technique is similar to cylindrical capacitor),$$ - V = - \frac{Q}{4\pi\epsilon_0} \int_a^b \frac{1}{r^2} dr $$ Thus,$$ V = \frac{Q}{4\pi\epsilon_0} \left(\frac{1}{a} -\frac{1}{b}\right) $$
And the capacitance is given by, $$ C = 4\pi\epsilon_0 \frac{ab}{b-a} $$
Thus we can find the capacitance for various shapes.
Our usual capacitor is just so small enough within the size of a finger, where the conducting material rolled down in a cylindrical shape with some non conducting or dielectric material in between them.
Let us discuss with mathematics,
The two plates have opposite charges, let say, "+Q" on the first plate and "-Q" on the second plate. So, the potential difference between these plates can be calculated as the work done required to bring a unit positive charge from -Q to +Q [from the definition of potential]
$$ \int_{V_-}^{V_+}\,dV = V = V_+ - V_- = - \int_{-}^{+} \vec{E}\cdot \vec{dl} \,\,\,...eq.(1)$$
We don't assume anything about the Shapes of the plates, so Electric field is just given by the definition as, $$ \vec{E} = \frac{1}{4\pi\epsilon_0} \int \rho \frac{\hat{r}}{r^2} d\tau $$ where $\rho$ is the volume charge density and integral is over the volume "V".
Electric field is proportional to both Potential "V" and the charge on the each capacitor "Q" and the ratio is some constant which is defined as the capacitance. $$ C = \frac{Q}{V} $$
Let us try to calculate the capacitance of some simple known shapes, where assumptions are easy to make.
Capacitance of Parallel Plate Capacitor
First one is the parallel plate capacitor, where Electric field is directed from positive charge to negative charge plate and assumed to be uniform in the direction. So, it can be taken out from the integral in equation (1) and integral sign is canceled by the negative sign and the equation becomes, $$ V = E \int_0^d \,dl = Ed $$ From our definition of Capacitance, $$ C = \frac{Q}{V} = {Q}{E d} $$ And, the Electric field in the region between parallel plate capacitors is given by, $$ E = \frac{\sigma}{\epsilon_0} $$ where $\sigma = \frac{Q}{A}$ is the surface charge density.
Then, $$ C = \frac{Q}{\frac{Qd}{A\epsilon_0}} = \frac{A\epsilon_0}{d} $$ which is determined only by the sizes, shapes, and separation distance of the two conductors.
Capacitance of Cylindrical capacitor
Second is the cylindrical capacitors, where inside is solid cylinder with positive charge with radius "a" and outside is hollow cylinder with bigger radius "b" (b>a). Let us take a point "r" in the "in between region a<r<b "
The Electric field in this region is given by making use of Gauss law in cylindrical symmetry as, $$ \vec{E} = \frac{\lambda}{2\pi r \epsilon_0} \hat{r}$$
Then potential difference is calculated by, $$ - V = - \int_a^b \vec{E} \cdot \vec{dr} $$ where $\vec{dr} $ is directed from "a" to "b" i.e. radially outwards. So, it becomes $$ V = \frac{\lambda}{2\pi \epsilon_0} \int_a^b \,\frac{1}{r} dr = \frac{\lambda }{2\pi\epsilon_0}\ln{(\frac{b}{a})}$$
where again the integral sign canceled by the negative sign(slightly different from the previous - Here we started with finding $-V = V_ - - V_+$ because the limits will be easy and all are measured from the centre - otherwise $\vec{dl} $ will be directed from -Q to +Q which will have limits 0 to "b-a" where logarithmic function not finite).
Thus we get the capacitance of the cylindrical capacitor as, $$ C = Q/V = \frac{\lambda{l} 2\pi\epsilon_0}{\lambda \ln{(\frac{b}{a})}} = \frac{2\pi\epsilon_0{l}}{\ln{(\frac{b}{a})}} $$
Capacitance of spherical capacitor
Finally, let us consider a spherical capacitor with two concentric spherical metal shells with radii a and b. Inner shell with +Q and outer shell with -Q charge.
In the same way, the Electric field in the region is given by, $$\vec{E} = \frac{Q}{4\pi\epsilon_0 r^2} \hat{r} $$ vector "r" points radially outwards.
Potential difference is calculated by, (integration technique is similar to cylindrical capacitor),$$ - V = - \frac{Q}{4\pi\epsilon_0} \int_a^b \frac{1}{r^2} dr $$ Thus,$$ V = \frac{Q}{4\pi\epsilon_0} \left(\frac{1}{a} -\frac{1}{b}\right) $$
And the capacitance is given by, $$ C = 4\pi\epsilon_0 \frac{ab}{b-a} $$
Thus we can find the capacitance for various shapes.
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