
What are the factors on which the electric field due to a point source charge depends?
Answer
407.4k+ views
Hint:we will give an answer to this question by first defining Coulomb’s law. We will then write the formula for the same. Later we will discuss the electric field in short and write a formula for the same. After writing the formula we will compare the terms to see what all the electric field depends on.
Formula used:
\[\left| F \right| = k\dfrac{{\left| {{q_1} \times {q_2}} \right|}}{{{r^2}}}\]
Where $F$ = force, $q_1$ and $q_2$ are 2 point charges and r is the distance between the 2 charges and $k$ is proportionality constant. \[k\approx \;8.988 \times {10^9}\;N \cdot {m^2} \cdot {C^{ - 2}}\].
\[F = q \times E\]
Complete step by step answer:
Let us look at Coulomb’s law: the law states that the magnitude of the electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them.
\[\left| F \right| = k\dfrac{{\left| {{q_1} \times {q_2}} \right|}}{{{r^2}}}\]
Electric field is given by \[E = \dfrac{F}{q}\]
The force is directed in the direction of the straight line that connects the two charges. The electrostatic force between two charges is repulsive if their signs are the same; if their signs are different, the force is attractive.
Let us look at what an electric field is: it is a region around a charged particle or object within which a force would be exerted on other charged particles or objects.
Hence \[E = k\dfrac{{\dfrac{{\left| {q \times q} \right|}}{{{r^2}}}}}{q}\]
i.e. \[E = k\dfrac{q}{{{r^2}}}\]
Hence clearly the electric field depends on magnitude of charge, distance between the point charges and medium between the charge and the point.
Note: Do not forget to mention the medium between the charge and the point.Looking at the formula we might just write that it is dependent on charge and distance between the charges. But it also depends on the type of medium in between them. The best example would be when dielectric is inserted.
Formula used:
\[\left| F \right| = k\dfrac{{\left| {{q_1} \times {q_2}} \right|}}{{{r^2}}}\]
Where $F$ = force, $q_1$ and $q_2$ are 2 point charges and r is the distance between the 2 charges and $k$ is proportionality constant. \[k\approx \;8.988 \times {10^9}\;N \cdot {m^2} \cdot {C^{ - 2}}\].
\[F = q \times E\]
Complete step by step answer:
Let us look at Coulomb’s law: the law states that the magnitude of the electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them.
\[\left| F \right| = k\dfrac{{\left| {{q_1} \times {q_2}} \right|}}{{{r^2}}}\]
Electric field is given by \[E = \dfrac{F}{q}\]
The force is directed in the direction of the straight line that connects the two charges. The electrostatic force between two charges is repulsive if their signs are the same; if their signs are different, the force is attractive.
Let us look at what an electric field is: it is a region around a charged particle or object within which a force would be exerted on other charged particles or objects.
Hence \[E = k\dfrac{{\dfrac{{\left| {q \times q} \right|}}{{{r^2}}}}}{q}\]
i.e. \[E = k\dfrac{q}{{{r^2}}}\]
Hence clearly the electric field depends on magnitude of charge, distance between the point charges and medium between the charge and the point.
Note: Do not forget to mention the medium between the charge and the point.Looking at the formula we might just write that it is dependent on charge and distance between the charges. But it also depends on the type of medium in between them. The best example would be when dielectric is inserted.
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