Two metal plates having charge $ + Q$ and $ - Q$ face each other at a certain distance between them. If the plates are now dipped in a kerosene oil tank, the electric fields between the plates will
A) Become zero
B) Increase
C) Decrease
D) Remains zero
Answer
299.7k+ views
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.
Recently Updated Pages
If the magnetizing field on a ferromagnetic material class 12 physics JEE_Main

A point charge is placed at the corner of a cube The class 12 physics JEE_Main

What changes occur if the monochromatic light used class 12 physics JEE_Main

The unit of specific conductance is A Ohm B Ohmmetre class 12 physics JEE_Main

Which lens is used in magnifying glass A Concave lens class 12 physics JEE_Main

Sir C V Raman won the Nobel Prize in which year A 1928 class 12 physics JEE_Main

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Understanding the Electric Field of a Uniformly Charged Ring

Electron Gain Enthalpy and Electron Affinity Explained

Derivation of Equation of Trajectory Explained for Students

Understanding Atomic Structure for Beginners

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
CBSE Class 12 Physics Question Paper 2026: Download SET-wise PDF with Answer Key & Analysis

JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

Effective Nuclear Charge for JEE

What Are Current and Potential Difference in Electricity?

Understanding Uniform Acceleration in Physics

Hybridisation in Chemistry – Concept, Types & Applications

