Answer
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Hint: To solve this type of question we use the relationship between the electric field and the dielectric constant. We can state that the electric field inside the dielectric constant (E) is directly proportional to the surface charge density given as $\sigma$ and inversely proportional to the dielectric constant $k$ and permittivity of the free space $\varepsilon_0$.
Formula used:
The electric field inside the dielectric $E = \dfrac{\sigma }{{K{\varepsilon_0}}}$; Here $k$ is dielectric constant, $\sigma$ is the surface charge density and $\varepsilon_0$ is the permittivity of the free space.
Complete answer:
Let us first write the information which is given in the question. It is given as follows,
$E =3 \times {10^4}V/m$,
$K = 2.2$ and
The distance between the plates = 5mm.
We know the value of ${\varepsilon _0} = 8.85 \times {10^{ - 12}}{C^2}{N^{ - 1}}{m^{ - 2}}$
Let us use the formula $E = \dfrac{\sigma }{{K{\varepsilon _0}}}$ and substitute the values which are given in the problem, we get
$\Rightarrow 3 \times {10^4} = \dfrac{\sigma }{{2.2 \times 8.85 \times {{10}^{ - 12}}}}$
Let us simplify it and find the value of surface charge density, we get,
$\Rightarrow \sigma = 3 \times 2.2 \times 8.85 \times {10^{ - 8}} \Rightarrow \sigma = 58.41 \times {10^{ - 8}}C{m^{ - 2}}$
We can also write the value of σ as below, we get
$\Rightarrow \sigma = 5.841 \times {10^{ - 7}}C{m^{ - 2}} \sim 6 \times {10^{ - 7}}C{m^{ - 2}}$
$\therefore $The charge density of the positive plate “$6 \times {10^{ - 7}}C{m^{ - 2}}$.” So, Option (A) is the correct option.
Note:
A device which stores the electrical energy is called a capacitor. It is a passive device with electrical components.
It has two conductors which are separated by a non-conductive region. The non-conductive area can be a vacuum or dielectric material.
The parallel plates in the capacitor store a finite amount of energy before breakdown.
Capacitors are used as energy storage, digital memory, pulsed power and weapons, and power conditioning, etc.
Dielectrics are electrically insulated and polarize on the application of the electric field. When a dielectric material is placed in an electric field, electric charges do not flow through the material as they do in an electrical conductor but only slightly shift from their average equilibrium positions causing dielectric polarization.
Formula used:
The electric field inside the dielectric $E = \dfrac{\sigma }{{K{\varepsilon_0}}}$; Here $k$ is dielectric constant, $\sigma$ is the surface charge density and $\varepsilon_0$ is the permittivity of the free space.
Complete answer:
Let us first write the information which is given in the question. It is given as follows,
$E =3 \times {10^4}V/m$,
$K = 2.2$ and
The distance between the plates = 5mm.
We know the value of ${\varepsilon _0} = 8.85 \times {10^{ - 12}}{C^2}{N^{ - 1}}{m^{ - 2}}$
Let us use the formula $E = \dfrac{\sigma }{{K{\varepsilon _0}}}$ and substitute the values which are given in the problem, we get
$\Rightarrow 3 \times {10^4} = \dfrac{\sigma }{{2.2 \times 8.85 \times {{10}^{ - 12}}}}$
Let us simplify it and find the value of surface charge density, we get,
$\Rightarrow \sigma = 3 \times 2.2 \times 8.85 \times {10^{ - 8}} \Rightarrow \sigma = 58.41 \times {10^{ - 8}}C{m^{ - 2}}$
We can also write the value of σ as below, we get
$\Rightarrow \sigma = 5.841 \times {10^{ - 7}}C{m^{ - 2}} \sim 6 \times {10^{ - 7}}C{m^{ - 2}}$
$\therefore $The charge density of the positive plate “$6 \times {10^{ - 7}}C{m^{ - 2}}$.” So, Option (A) is the correct option.
Note:
A device which stores the electrical energy is called a capacitor. It is a passive device with electrical components.
It has two conductors which are separated by a non-conductive region. The non-conductive area can be a vacuum or dielectric material.
The parallel plates in the capacitor store a finite amount of energy before breakdown.
Capacitors are used as energy storage, digital memory, pulsed power and weapons, and power conditioning, etc.
Dielectrics are electrically insulated and polarize on the application of the electric field. When a dielectric material is placed in an electric field, electric charges do not flow through the material as they do in an electrical conductor but only slightly shift from their average equilibrium positions causing dielectric polarization.
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