Questions & Answers
Question
AnswersThe value of constant \[k\] in coulomb’s law is:
\[\begin{align}
& A)9\times {{10}^{3}} \\
& B)9\times {{10}^{5}} \\
& C)9\times {{10}^{7}} \\
& D)9\times {{10}^{9}} \\
\end{align}\]
Answer
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Hint: We will begin with coulomb’s law which gives us the magnitude of electrostatic force between two point charges separated by a distance. Then we will find the value of constant \[k\] from the equation by using the value for permittivity of free space (\[{{\varepsilon }_{0}}\]) which is a physical constant used in electromagnetism which represents the capability of vacuum to permit electric fields.
Formula used:
\[F=\dfrac{1}{4\pi{{\varepsilon}_{0}}}\times\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}=k\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\]
Complete step by step answer:
We know coulomb’s law gives the expression for magnitude of electrostatic force between two charges at some distance as,
\[F=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\times \dfrac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\]
Where, \[{{q}_{1}}\] and \[{{q}_{2}}\] are the two charges.
\[r\] is the distance between two charges.
\[{{\varepsilon }_{0}}\] is a permittivity of vacuum.
Permittivity of vacuum (\[{{\varepsilon }_{0}}\]) is the ability of vacuum to permit electric field lines. It is always constant. \[{{\varepsilon }_{0}}\] has a value of,
\[{{\varepsilon }_{0}}=8.854\times {{10}^{-12}}F/m\]
Now, we will take the constant part of coulomb’s equation as \[k\]. It is given as,
\[k=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\]
Here, \[\pi \] is taken as \[\pi =3.14\].
Then, the value of \[k\] will be,
\[k=\dfrac{1}{4\times 3.14\times 8.854\times {{10}^{-12}}}\approx 9\times {{10}^{9}}\]
So, the value of constant \[k\] is coulomb’s equation is found as \[k=9\times {{10}^{9}}\].
Hence, the correct answer is option D.
Note:
We have a great use of remembering this constant because we can directly substitute this for the constant part in the equation. Also this constant is used for finding electric field intensity and electric potential energy. We must know that this constant depends on the medium in which electric field lines are present. This vale for the constant is not applicable in a medium other than vacuum. We will need to substitute the permittivity of that medium to find the value of this constant in those mediums.
Formula used:
\[F=\dfrac{1}{4\pi{{\varepsilon}_{0}}}\times\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}=k\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\]
Complete step by step answer:
We know coulomb’s law gives the expression for magnitude of electrostatic force between two charges at some distance as,
\[F=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\times \dfrac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\]
Where, \[{{q}_{1}}\] and \[{{q}_{2}}\] are the two charges.
\[r\] is the distance between two charges.
\[{{\varepsilon }_{0}}\] is a permittivity of vacuum.
Permittivity of vacuum (\[{{\varepsilon }_{0}}\]) is the ability of vacuum to permit electric field lines. It is always constant. \[{{\varepsilon }_{0}}\] has a value of,
\[{{\varepsilon }_{0}}=8.854\times {{10}^{-12}}F/m\]
Now, we will take the constant part of coulomb’s equation as \[k\]. It is given as,
\[k=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\]
Here, \[\pi \] is taken as \[\pi =3.14\].
Then, the value of \[k\] will be,
\[k=\dfrac{1}{4\times 3.14\times 8.854\times {{10}^{-12}}}\approx 9\times {{10}^{9}}\]
So, the value of constant \[k\] is coulomb’s equation is found as \[k=9\times {{10}^{9}}\].
Hence, the correct answer is option D.
Note:
We have a great use of remembering this constant because we can directly substitute this for the constant part in the equation. Also this constant is used for finding electric field intensity and electric potential energy. We must know that this constant depends on the medium in which electric field lines are present. This vale for the constant is not applicable in a medium other than vacuum. We will need to substitute the permittivity of that medium to find the value of this constant in those mediums.