Gurmeet has to reach his friend house which is 36 km away from his house in 45 minutes. Should he travel by the city bus, which has an average speed o … f 30km/h or the train which has an average speed of 50 km/h?
i already told u science is all about facts. if u think science is borring then u learning it from wring teacher. u know what I got 99 only in physics … and 98 in chemistry in my board examination.
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yjvxocchxu baddies With law enforcements agents in hot pursuit, and police cars blocking their way, E knows the squad is doomed and closes his eyes. S … uddenly, his X surges skyward over the cars, and he and his friends find themselves flying high over the rooftops. With the fiery orb of the setting sun as the backdrop, the visitor propels them towards the woods so he can go home. For the answer, figure out X and E. The hash verification is X+E in lowercase, but for later, all you really need yjvxocchxu
i am orange, i am white i am dark as i am dyed take the shape you wish for your eyes. i am dead solid yet rumbling alive. who am i?
i am orange, i am white i am dark as i am dyed take the shape you wish for your eyes. i am dead solid yet rumbling alive. who am i?
what does an electric current mean?
mention the three way by which the over loading occurs what is short circuiting ?
- What is series LCR resonant circuit? State conditions for series resonance. Obtain an expression for resonant frequency.
4. integrate ----- x sin 2x dx
Answer
Hint: We are given two point charges kept in air and the distance of separation between them. It is said the same charges are placed in oil and the force between the charges remains the same. We are also given the relative permittivity of oil. Since it is said that the force between the charges are equal in the two mediums we can equate the force experienced in two mediums. By substituting the known values and eliminating the common terms we get the solution.
Formula used: Force experienced between two charges, $F=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$
Complete step by step answer:
In the question it is said that two point charges ${{q}_{1}}$ and ${{q}_{2}}$ are placed in air and the distance of separation between them is 50 m.Let ‘${{r}_{1}}$’ be the distance of separation in air, then${{r}_{1}}=50m$ It is said that the same point charges are placed in oil.The relative permittivity of oil is given to us, ${{\varepsilon }_{1}}=5$, were ‘${{\varepsilon }_{1}}$’ is the relative permittivity of oil. We know that the force between two charges is given by the equation,$F=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$, were’ ‘F’ is force, ‘$\dfrac{1}{4\pi {{\varepsilon }_{0}}}$’ is a constant, ‘${{q}_{1}}$’ and ‘${{q}_{2}}$’ are the two charges, ‘${{\varepsilon }_{0}}$’ is the relative permittivity of the medium and ‘$r$’ is the separating distance between the two charges.Now let us consider the case when the charges are placed in the air. The force between the two charges,${{F}_{1}}=\dfrac{1}{4\pi \varepsilon }\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}_{1}}^{2}}$ Here the medium is air; we know that the relative permittivity of air is 1.Therefore, we have$\Rightarrow {{F}_{1}}=\dfrac{1}{4\pi }\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}_{1}}^{2}}$Now let us consider the case when the charges are placed in oil.The force between the two charges will be,${{F}_{2}}=\dfrac{1}{4\pi {{\varepsilon }_{1}}}\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}_{2}}^{2}}$ Since relative permittivity of oil is given, ${{\varepsilon }_{1}}=5$ we get$\Rightarrow {{F}_{2}}=\dfrac{1}{4\pi \times \left( 5 \right)}\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}_{2}}^{2}}$ In the question it is said that the force between the two charges is the when in both mediums.Hence we can equate ${{F}_{1}}$ and ${{F}_{2}}$ \[\Rightarrow {{F}_{1}}={{F}_{2}}\] \[\Rightarrow \dfrac{1}{4\pi }\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}_{1}}^{2}}=\dfrac{1}{4\pi \times \left( 5 \right)}\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}_{2}}^{2}}\] \[\Rightarrow \dfrac{1}{{{r}_{1}}^{2}}=\dfrac{1}{5\times {{r}_{2}}^{2}}\] \[\Rightarrow 5\times {{r}_{2}}^{2}={{r}_{1}}^{2}\] \[\Rightarrow {{r}_{2}}^{2}=\dfrac{{{r}_{1}}^{2}}{5}\] \[\Rightarrow {{r}_{2}}=\sqrt{\dfrac{{{r}_{1}}^{2}}{5}}\] We know that ${{r}_{1}}=50m$, by substituting this in the above equation, we get\[\Rightarrow {{r}_{2}}=\sqrt{\dfrac{{{50}^{2}}}{5}}\] \[\Rightarrow {{r}_{2}}=\sqrt{\dfrac{2500}{5}}\] \[\Rightarrow {{r}_{2}}=\sqrt{500}\] \[\Rightarrow {{r}_{2}}=22.36m\] Therefore the distance of separation between the two charges in oil is 22.36 m.So, the correct answer is “Option B”.
Note: The force experienced between two charges is given by Coulomb’s law.
According to Coulomb’s law the electrical force between two charges separated by a distance is directly proportional to the product of the magnitude of the two charges and is inversely proportional to the square separating distance between the charges.This proportionality is equated by the constant $\dfrac{1}{4\pi {{\varepsilon }_{0}}}$ which is $=9\times {{10}^{9}}N{{m}^{2}}/{{C}^{2}}$