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Step-by-Step Explanation
Step 1: Understand the Scenario
We have two parallel metal plates carrying charges +Q and βQ. In vacuum (or air), there is a certain electric field between the plates. The question is about what happens to this electric field when the plates are dipped in kerosene oil (a dielectric medium).
Step 2: Electric Field in Vacuum
When two oppositely charged parallel plates are placed facing each other in vacuum or air, the electric field between them is given by
$$E = \frac{\sigma}{\epsilon_0},$$
where
Ο is the surface charge density on the plates (charge per unit area), and
Ξ΅β is the permittivity of free space.
Step 3: Electric Field in a Dielectric Medium
When these plates are immersed in a medium with a dielectric constant K (K > 1 for most dielectric materials, including kerosene oil), the electric field between the plates becomes
$$E' = \frac{\sigma}{K \,\epsilon_0}.$$
Since K > 1, it follows that
$$E' = \frac{E}{K} \quad \Rightarrow \quad E' < E.$$
Thus, the presence of the dielectric reduces the electric field strength.
Step 4: Comparison of Electric Fields
Comparing the electric field in vacuum (E) with that in kerosene oil (E'):
β’ In vacuum (or air): $E = \frac{\sigma}{\epsilon_0}$
β’ In kerosene oil: $E' = \frac{\sigma}{K \,\epsilon_0}$
Because K > 1, we have $E' < E$.
Step 5: Conclusion
The electric field between the charged plates decreases when the plates are dipped in kerosene oil. Hence, the correct choice is that the electric field βdecreases.β
