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Question: In a parallel plate capacitor with air between the plates, each plate has an area of \[6 \times {10^...

In a parallel plate capacitor with air between the plates, each plate has an area of 6×10−3m26 \times {10^{ - 3}}{m^2} and the distance between the plates is 3mm3mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100V100V supply, what is the charge on each plate of the capacitor?

Explanation

Solution

Hint : Find the capacitance of the capacitor using the given data. It is given as the ratio between the product of area and medium, divided by the distance between the plates. Using the capacitance value, find the charge of the capacitor when connected to supply, by just multiplying with the supply voltage value.

Complete Step by Step Answer:
In the given data, it is mentioned that there is a parallel plate capacitor which uses air as a di-electric medium. Each plate has an area of 6×10−3m26 \times {10^{ - 3}}{m^2}and is said to be connected with a gap of 3mm3mmfrom each plate.
Capacitance is defined as the ability of the device to hold up charge, and discharge the same amount of power without loss. Now, capacitance for a parallel plate capacitor can be calculated using the formula,
⇒C=Aε0d\Rightarrow C = \dfrac{{A{\varepsilon _0}}}{d}, Where A is the area of the parallel plate, ε0{\varepsilon _0}is the electric permittivity of free space and d is the distance of separation between the plates.
Substituting the values, we get
⇒C=6×10−3×8.85×10−120.003m\Rightarrow C = \dfrac{{6 \times {{10}^{ - 3}} \times 8.85 \times {{10}^{ - 12}}}}{{0.003m}}
Further simplifying, we get the value as ,
⇒C=53.1×10−150.003m\Rightarrow C = \dfrac{{53.1 \times {{10}^{ - 15}}}}{{0.003m}}
Further simplifying, we get,
⇒C=53.1×10−150.003m=17.7×10−12F\Rightarrow C = \dfrac{{53.1 \times {{10}^{ - 15}}}}{{0.003m}} = 17.7 \times {10^{ - 12}}F
Now, when the capacitor is connected to a source of 100V100V, it begins to obtain charge. Now, the charge flowing through the capacitor is given as the product of its capacitance and the source voltage. Mathematically,
⇒q=CV\Rightarrow q = CV
Substituting Capacitance and Voltage , we get,
⇒q=CV=17.7×10−12×100\Rightarrow q = CV = 17.7 \times {10^{ - 12}} \times 100
⇒q=17.7×10−10C\Rightarrow q = 17.7 \times {10^{ - 10}}C
Thus, the capacitance and the charge flowing through it are identified using the given data.
Note: Permittivity of free space , also termed as vacuum permittivity is the constant value that is generally considered , when air is taken as transmission medium. It mainly represents the capacity of vacuum space to allow and conduct electric fields.