If two charges of 1 coulomb each are placed 1 km apart, the force between them will be
(1)\[9\times {{10}^{3}}\]N (2)\[9\times {{10}^{-3}}\] (3)\[1.1\times {{10}^{-4}}\] (4)\[{{10}^{-6}}\]
Answer
587.9k+ views
Hint: We will use the formula of coulomb's electrostatic force between two charges. By substituting the values of k, charge and the distance between them we can find the value of Force between the two charges.
Formula: $F=k\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$
Complete answer:
We know that electrostatic force acts between two stationary electrically charged particles separated by a distance ‘r’. This force is also called coulomb's force. It is given by the formula
$F=k\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$ ---------(1)
Where k is the coulomb's constant, ${{q}_{1}}$and ${{q}_{2}}$is the charge in coulombs.
In the question, it is given that ${{q}_{1}}={{q}_{2}}=1C$
And the charges are placed 1km apart, so $r=1km=1000m$
The value of coulomb’s constant $k=9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}}$
Substituting above values in equation (1), we get
$\begin{align}
& \Rightarrow 9\times {{10}^{9}}\times \dfrac{1\times 1}{{{(1000)}^{2}}}N \\
& \Rightarrow 9\times {{10}^{9}}\times {{10}^{-6}}N \\
& \Rightarrow 9\times {{10}^{3}}N \\
& \therefore F=9\times {{10}^{3}}N \\
\end{align}$
So, If two charges of 1 coulomb are placed 1 km apart, the force between them will be$9\times {{10}^{3}}N$.
So, Option (1) is correct.
Additional information:
Electrostatic force is an attractive or repulsive force which acts between stationary electrically charged particles. The electric force between a stationary charged body is conventionally known as the electrostatic force or Coulomb’s force.
1 Coulomb charge is equal to the amount of charge from a current of one ampere flowing for one second. It can also be defined as One coulomb is equal to the charge on $6.241\times {{10}^{18}}$ protons. The charge on 1 proton is $1.6\times {{10}^{-19}}$ C.
Note:
In the formula for electrostatic force, ‘k’ is the proportionality constant also known as coulomb’s constant.
\[\begin{align}
& k=\dfrac{1}{4\pi {{\xi }_{0}}}=9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}} \\
& {{\xi }_{0}}=8.85\times {{10}^{-12}}{{C}^{2}}/N{{m}^{2}} \\
\end{align}\]
${{\xi }_{0}}$ is known as the permittivity of free space.
One must remember all these values while solving questions of electrostatics.
Formula: $F=k\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$
Complete answer:
We know that electrostatic force acts between two stationary electrically charged particles separated by a distance ‘r’. This force is also called coulomb's force. It is given by the formula
$F=k\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$ ---------(1)
Where k is the coulomb's constant, ${{q}_{1}}$and ${{q}_{2}}$is the charge in coulombs.
In the question, it is given that ${{q}_{1}}={{q}_{2}}=1C$
And the charges are placed 1km apart, so $r=1km=1000m$
The value of coulomb’s constant $k=9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}}$
Substituting above values in equation (1), we get
$\begin{align}
& \Rightarrow 9\times {{10}^{9}}\times \dfrac{1\times 1}{{{(1000)}^{2}}}N \\
& \Rightarrow 9\times {{10}^{9}}\times {{10}^{-6}}N \\
& \Rightarrow 9\times {{10}^{3}}N \\
& \therefore F=9\times {{10}^{3}}N \\
\end{align}$
So, If two charges of 1 coulomb are placed 1 km apart, the force between them will be$9\times {{10}^{3}}N$.
So, Option (1) is correct.
Additional information:
Electrostatic force is an attractive or repulsive force which acts between stationary electrically charged particles. The electric force between a stationary charged body is conventionally known as the electrostatic force or Coulomb’s force.
1 Coulomb charge is equal to the amount of charge from a current of one ampere flowing for one second. It can also be defined as One coulomb is equal to the charge on $6.241\times {{10}^{18}}$ protons. The charge on 1 proton is $1.6\times {{10}^{-19}}$ C.
Note:
In the formula for electrostatic force, ‘k’ is the proportionality constant also known as coulomb’s constant.
\[\begin{align}
& k=\dfrac{1}{4\pi {{\xi }_{0}}}=9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}} \\
& {{\xi }_{0}}=8.85\times {{10}^{-12}}{{C}^{2}}/N{{m}^{2}} \\
\end{align}\]
${{\xi }_{0}}$ is known as the permittivity of free space.
One must remember all these values while solving questions of electrostatics.
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

What do you mean by retardation What is its SI uni class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Difference between physical and chemical change class 11 chemistry CBSE

