
The dimension of permittivity $(\varepsilon _0)$ are ______. Take Q as the dimension of charge.
A) $[M^{-1} L^{-2} T^{-2} Q^{-2}]$
B) $[M^{-1} L^{-3} T^{2} Q^{2}]$
C) $[M^{-1} L^{2} T^{-3} Q^{-1}]$
D) $[M^{-1} L^{3} T^{-2} Q^{-2}]$
Answer
132.9k+ views
Hint: The dimensions of a physical quantity are the power raised by the basic units to obtain a unit of those quantities. For derived quantity represented by \[{Q = }{{{M}}^{{a}}}{{{L}}^{{b}}}{{{T}}^{{c}}}\], then \[{{{M}}^{{a}}}{{{L}}^{{b}}}{{{T}}^{{c}}}\] is called dimensional formula.
Complete solution:
${F = }\dfrac{{K}{{{Q}}^{{2}}}}{{{{r}}^{{2}}}}$
and dimension of $F = [M L T^{-2}]$
$\Rightarrow {ML}{{{T}}^{-2}}{ = }\dfrac{{K}{{{Q}}^{{2}}}}{{{{r}}^{{2}}}}$
$\Rightarrow K = [M L^3 T^{-2} Q^{-2}]$
Substituting value of K
$\Rightarrow \dfrac{1}{4 \pi \varepsilon _0}$
Hence, answer is option (B), $[M^{-1} L^{-3} T^2 Q^2]$.
Additional Information:
Permeability and dielectric constant are two terms that are central to capacitor technology. It is often talked about that capacitors will be used with different dielectrics. All common names for electrolytic capacitors, ceramic capacitors, paper, tantalum capacitors, and capacitors refer to the dielectric material used.
The dielectric components provide insulation between the capacitor plates and in addition it determines many of the properties of the capacitor. This capacitance is achieved in constant quantities, whether the temperature constant is polarized or not. These and many other properties are controlled by the dielectric material used - many properties are managed only by the dielectric constant.
Note: Vacuum refers to the minimum possible value of permittivity allowed. This is commonly known as free space or permitted electronic constant. Denoted by \[ \varepsilon _0\] and has the value \[8.85 \times {10}^{-12} Farad/meter\]. Disputes against the formation of electric field lines are also evident in the indictment. The permittivity of a dielectric is represented by the ratio of its absolute permittivity to the electronic constant and is usually given relative to the erase.
Complete solution:
${F = }\dfrac{{K}{{{Q}}^{{2}}}}{{{{r}}^{{2}}}}$
and dimension of $F = [M L T^{-2}]$
$\Rightarrow {ML}{{{T}}^{-2}}{ = }\dfrac{{K}{{{Q}}^{{2}}}}{{{{r}}^{{2}}}}$
$\Rightarrow K = [M L^3 T^{-2} Q^{-2}]$
Substituting value of K
$\Rightarrow \dfrac{1}{4 \pi \varepsilon _0}$
Hence, answer is option (B), $[M^{-1} L^{-3} T^2 Q^2]$.
Additional Information:
Permeability and dielectric constant are two terms that are central to capacitor technology. It is often talked about that capacitors will be used with different dielectrics. All common names for electrolytic capacitors, ceramic capacitors, paper, tantalum capacitors, and capacitors refer to the dielectric material used.
The dielectric components provide insulation between the capacitor plates and in addition it determines many of the properties of the capacitor. This capacitance is achieved in constant quantities, whether the temperature constant is polarized or not. These and many other properties are controlled by the dielectric material used - many properties are managed only by the dielectric constant.
Note: Vacuum refers to the minimum possible value of permittivity allowed. This is commonly known as free space or permitted electronic constant. Denoted by \[ \varepsilon _0\] and has the value \[8.85 \times {10}^{-12} Farad/meter\]. Disputes against the formation of electric field lines are also evident in the indictment. The permittivity of a dielectric is represented by the ratio of its absolute permittivity to the electronic constant and is usually given relative to the erase.
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