Class 12PhysicsElectric Potential and CapacitanceConductor, Electrostatic Shielding, Induced Charge and Charge Redistribution on conductor
MediumWhat is earthing or grounding in household electrical circuits?
(N/A) When we bring a charged body in contact with the earth,all the excess charge on the body disappears by causing a momentary current to pass to the ground through the connecting conductor. This process of sharing the charges with the earth is called grounding or earthing.
$A$ thick metal plate is buried deep into the earth and thick wires are drawn from this plate; these are used in buildings for the purpose of earthing near the mains supply.
The electric wiring in our houses has three wires: $live$,$neutral$,and $earth$.
The first two carry electric current from the power station and the third is earthed by connecting it to the buried metal plate.
Metallic bodies of electric appliances such as electric iron,refrigerator,and $TV$ are connected to the earth wire. When any fault occurs or a live wire touches the metallic body,the charge flows to the earth without damaging the appliance and without causing any injury to humans.
Earthing provides a safety measure for the human body,electrical circuits,and appliances.
$A$ thick metal plate is buried deep into the earth and thick wires are drawn from this plate; these are used in buildings for the purpose of earthing near the mains supply.
The electric wiring in our houses has three wires: $live$,$neutral$,and $earth$.
The first two carry electric current from the power station and the third is earthed by connecting it to the buried metal plate.
Metallic bodies of electric appliances such as electric iron,refrigerator,and $TV$ are connected to the earth wire. When any fault occurs or a live wire touches the metallic body,the charge flows to the earth without damaging the appliance and without causing any injury to humans.
Earthing provides a safety measure for the human body,electrical circuits,and appliances.
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Similar Questions
Which of the following statements are correct?
$A$. Inside the conductor, the electrostatic field is zero.
$B$. Electric field at the surface of a charged conductor does not depend on its surface charge density.
$C$. The interior of a charged conductor can have no excess charge in the static situation.
$D$. At the surface of a charged conductor, the electrostatic field must be normal to the surface at every point.
$E$. The electrostatic potential is zero everywhere inside a charged conductor.
Choose the correct answer from the options given below:
$A$. Inside the conductor, the electrostatic field is zero.
$B$. Electric field at the surface of a charged conductor does not depend on its surface charge density.
$C$. The interior of a charged conductor can have no excess charge in the static situation.
$D$. At the surface of a charged conductor, the electrostatic field must be normal to the surface at every point.
$E$. The electrostatic potential is zero everywhere inside a charged conductor.
Choose the correct answer from the options given below:
MediumNEET 2026
View SolutionTwo isolated metallic solid spheres of radii $R$ and $2R$ are charged such that both have the same charge density $\sigma$. The spheres are then connected by a thin conducting wire. If the new charge density of the bigger sphere is $\sigma^{\prime}$,then the ratio $\frac{\sigma^{\prime}}{\sigma}$ is:
Two spheres $A$ and $B$ are kept very near to each other. $A$ is negatively charged and $B$ is earthed. The true statement is:
$(A)$ Charge on $B$ is zero
$(B)$ Potential at $B$ is zero
$(C)$ Charge is uniformly distributed on $A$
$(D)$ Charge is non-uniformly distributed on $A$
$(A)$ Charge on $B$ is zero
$(B)$ Potential at $B$ is zero
$(C)$ Charge is uniformly distributed on $A$
$(D)$ Charge is non-uniformly distributed on $A$
Medium
View SolutionTwo insulated charged conducting spheres of radii $20\,cm$ and $15\,cm$ respectively and having an equal charge of $10\,C$ are connected by a copper wire and then they are separated. Then
Easy
View SolutionTwo spheres of radius $a$ and $b$ respectively are charged and joined by a wire. The ratio of the electric field at the surfaces of the spheres is
Difficult
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