
What is the unit of electric field intensity?
${\text{A}}{\text{.}}$ Newton per metre
${\text{B}}{\text{.}}$ Coulomb per newton
${\text{C}}{\text{.}}$ Newton per coulomb
${\text{D}}{\text{.}}$ Joule per newton
Answer
517.8k+ views
Hint- Here, we will proceed by defining the physical term electric field intensity. We will also write down the formula for calculating electric field intensity. Through this formula, we can easily know the S.I units of electric field intensity.
Complete step by step answer:
Electric charge is carried by the subatomic particles of an atom such as electrons and photons. The charge of an electron is about $1.6 \times {10^{ - 19}}$ coulombs. Every charged particle creates a space around it in which the effect of its electric force is felt. This space around the charged particles is known as the “Electric field“. Whenever a unit test charge is placed in this electric field it will experience the force emitted by the source particle. The amount of force experienced by a unit charged particle when it is placed in the electric field is known as Electric field intensity.
The formula for electric field intensity is given by
Electric field intensity = $\dfrac{{{\text{Force experienced}}}}{{{\text{Magnitude of the charge}}}}$ i.e., E = $\dfrac{{\text{F}}}{{\text{q}}}$
The S.I unit of electric field intensity is newton per coulomb.
Hence, option C is correct.
Note- Since, the S.I units of force is newton and the S.I unit of charge magnitude is coulomb that’s why the S.I units of electric field intensity E = $\dfrac{{\text{F}}}{{\text{q}}}$ will be newton per coulomb. Also, note that electric field intensity is a vector quantity that means it will possess both magnitude as well as direction.
Complete step by step answer:
Electric charge is carried by the subatomic particles of an atom such as electrons and photons. The charge of an electron is about $1.6 \times {10^{ - 19}}$ coulombs. Every charged particle creates a space around it in which the effect of its electric force is felt. This space around the charged particles is known as the “Electric field“. Whenever a unit test charge is placed in this electric field it will experience the force emitted by the source particle. The amount of force experienced by a unit charged particle when it is placed in the electric field is known as Electric field intensity.
The formula for electric field intensity is given by
Electric field intensity = $\dfrac{{{\text{Force experienced}}}}{{{\text{Magnitude of the charge}}}}$ i.e., E = $\dfrac{{\text{F}}}{{\text{q}}}$
The S.I unit of electric field intensity is newton per coulomb.
Hence, option C is correct.
Note- Since, the S.I units of force is newton and the S.I unit of charge magnitude is coulomb that’s why the S.I units of electric field intensity E = $\dfrac{{\text{F}}}{{\text{q}}}$ will be newton per coulomb. Also, note that electric field intensity is a vector quantity that means it will possess both magnitude as well as direction.
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