By Sunil Bhardwaj

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CAPACITANCE OF PARALLEL PLATE CAPACITOR

Consider a parallel plate capacitor. The plates are separated by distance $$d$$ and each plate has area $$A$$. The plate $$a$$ has charge $$+Q$$ and plate $$b$$ has charge $$-Q$$. Imagine a Gaussian surface which encloses the charge $$+Q$$ of the capacitor plate as shown in fig.

Take small area element $$\vec{da}$$ of Gaussian surface. The electric flux through this area element is $$d\phi_c=\vec{E}.\vec{da}$$ Where $$d\phi_c$$ is the electric flux and $$\vec{E}$$ is electric field.

To calculate The net flux through Gaussian surface, we need to integrate this equation as $$\phi_c=\oint{\vec{E}.\vec{da}}$$ As we know the multiple of two vectors is given as $$\vec{A}.\vec{B}\ =\ A.B\ \cos{\theta}$$ Where $$\theta$$ is the angle between the directions of two vectors. Similarly, above equation becomes. $$\phi_c=E. da. \cos0°$$ As the direction of $$\vec{E}$$ and $$\vec{da}$$ is same the angle is $$0°$$ $$\therefore\phi_c=\oint{E\ da\ }$$ $$\therefore\phi_c=E.A$$

. . . . . . . . . . . . . (1)

Where $$A$$ is the area of that part of the Gaussian surface through which electric flux is passing. But from Gauss's law electric flux is given as $$\phi_c=\frac{Q}{\varepsilon_0}$$

. . . . . . . . . . . . . (2)

Where $$Q$$ is the charge and $$\varepsilon_0$$ is the permittivity of the medium. Comparing equation (1) and (2) we get, $$E\ A=\frac{Q}{\varepsilon_0}$$ $$E=\frac{Q}{A\ \varepsilon_0}$$

. . . . . . . . . . . . . (3)

Now lets calculate the potential difference between plates of the capacitor. Potential difference between two points a and b is given by, $$V_b-V_a=-\int_{a}^{b}{\vec{E}.\vec{ds}}$$ Where $$\vec{E}$$ is the electric field and $$\vec{ds}$$ is the separation or distance between points $$a$$ and $$b$$. As previously done lets multiply two vectors again. $$V_b-V_a=-\int_{a}^{b}{E\ ds\cos{\theta}}$$ The angle between $$ds$$ and $$E$$ is zero because both have same direction $$V_b-V_a=-\int_{a}^{b}{E\ ds}$$ For absolute potential between plates, put $$V_b=0$$. $$V=\int_{a}^{b}{E\ ds}$$ The potential between plates having separation $$d$$ is $$V=\int_{0}^{d}{E\ ds}$$ $$V=E.d$$ Where $$d$$ is the total separation between two plates. Lets add the value of $$E$$ from equation (3) $$V=\frac{Q}{A\ \varepsilon_0}d$$ On rearranging the equation $$\frac{Q}{V}=\frac{A\ \varepsilon_0}{d}$$ $$\frac{Q}{V}$$ is the capacitance $$C$$ of the parallel plate capacitor. $$\therefore C=\frac{A\ \varepsilon_0}{d}$$ The capacitance of a parallel plate capacitor depends upon area of plates, separation between plates. In addition, the dielectric medium inserted between plates also affect capacitance of parallel plate capacitor.

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