Answer
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Hint: In this solution, we will first calculate the formula for electric fields between two metal plates. Then we will determine its dependence on permittivity which will be affected when the metal plates are dipped in kerosene.
Formula used: In this solution, we will use the following formulae:
Electric field between two charged plates $E = \dfrac{\sigma }{{{\varepsilon _0}}}$ where $\sigma $ is the surface charge density and ${\varepsilon _0}$ is the permittivity in a vacuum.
Complete step by step answer:
We’ve been given that two opposite electrically charged plates are placed parallel to each other. We know that the electric field due to one charged plate is proportional to the charge density on the plate. This can be mathematically written as:
$E = \dfrac{\sigma }{{2{\varepsilon _0}}}$
Since the plates are charged opposite, their electric fields will be in the same direction, and hence the net electric field due to 2 plates add up. So the net electric field will be:
${E_{net}} = \dfrac{\sigma }{{2{\varepsilon _0}}} + \dfrac{\sigma }{{2{\varepsilon _0}}}$
$ \Rightarrow {E_{net}} = \dfrac{\sigma }{{{\varepsilon _0}}}$
As we can see in the above equation, the electric field is inversely proportional to the permittivity of the medium in which the two plates are placed in.
Now when the two plates are dipped in kerosene, the permittivity will increase. From the above relation, we can see that as the permittivity increases, the electric field will decrease.
So, option (C) is correct.
Note: We don’t have to know the permittivity of the kerosene to answer this question. Unless mentioned otherwise, we can assume the medium between the plates in the prior case to be a vacuum, and hence the permittivity of any medium will be higher than vacuum so the electric field will be lower for any medium.
Formula used: In this solution, we will use the following formulae:
Electric field between two charged plates $E = \dfrac{\sigma }{{{\varepsilon _0}}}$ where $\sigma $ is the surface charge density and ${\varepsilon _0}$ is the permittivity in a vacuum.
Complete step by step answer:
We’ve been given that two opposite electrically charged plates are placed parallel to each other. We know that the electric field due to one charged plate is proportional to the charge density on the plate. This can be mathematically written as:
$E = \dfrac{\sigma }{{2{\varepsilon _0}}}$
Since the plates are charged opposite, their electric fields will be in the same direction, and hence the net electric field due to 2 plates add up. So the net electric field will be:
${E_{net}} = \dfrac{\sigma }{{2{\varepsilon _0}}} + \dfrac{\sigma }{{2{\varepsilon _0}}}$
$ \Rightarrow {E_{net}} = \dfrac{\sigma }{{{\varepsilon _0}}}$
As we can see in the above equation, the electric field is inversely proportional to the permittivity of the medium in which the two plates are placed in.
Now when the two plates are dipped in kerosene, the permittivity will increase. From the above relation, we can see that as the permittivity increases, the electric field will decrease.
So, option (C) is correct.
Note: We don’t have to know the permittivity of the kerosene to answer this question. Unless mentioned otherwise, we can assume the medium between the plates in the prior case to be a vacuum, and hence the permittivity of any medium will be higher than vacuum so the electric field will be lower for any medium.
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