The force between two charges $0.06m$ apart is $5N$ . If each charge is moved towards the other by $0.01m$ , then the force between them will become.
A.$7.20N$
B.$11.25N$
C.$22.50N$
D.$45.00N$
Answer
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Hint: To solve this problem firstly we will learn about the electrostatics, Coulomb’s law and the concept behind these terms. We will collect the given information from the above question and apply the required formula, and by this we can approach the answer to our question easily.
Complete answer:
Electrostatics: Electrostatics branch of physics deals with the electric effect of static charge is called electrostatics. The effect of stationary or slow- moving electric charges. Electrostatic cause from the force that electric charge exerts on each other and are described by Coulomb’s law.
Coulomb’s Law: It begins with the magnitude of the electrostatic force between two-point charges $q\,\,and\,Q$ . It is convenient to label one of these charges, $q$ as a test charge and $Q$ a source charge. As we develop the theory, more source charges will be beaded. If $r$ is the distance between two charges, then the force of electrostatic formula is:
$
F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{qQ}}{{{r^2}}} = {k_e}\dfrac{{qQ}}{{{r^2}}} \\
F = k\dfrac{{{q_1}{q_2}}}{{{d^2}}} \\
$
Force between $2$ charges $0.06m$ apart is $5N$
We know, $F = \dfrac{{K{q_1}{q_2}}}{{{r^2}}}$
$S = \dfrac{{K{q_1}{q_2}}}{{{{\left( {0.06} \right)}^2}}}$
$K{q_1}{q_2} = 0.018..................\left( 1 \right)$
${F^1} = \dfrac{{K{q_1}{q_2}}}{{{{\left[ {0.06 - 2\left( {0.01} \right)} \right]}^2}}}$ charge is moved with regard to the other by $0.04m$
${F^1} = \dfrac{{0.018}}{{{{\left( {0.04} \right)}^2}}}$
${F^1} = 11.25N$
Force will become $11.25$ newtons.
Hence, the correct option is B $11.25N$
Note: Coulomb’s law of physics is the incident that measures the amount of force between two stationaries, electrically charged particles. The electric force between charge bodies at rest is normally called electrostatic force. There are many examples of electrostatic events, the attraction of the plastic wrapped on hand after removing it from a package, the attraction of paper to a charged scale, the apparently unconstrained explosion of grain silos. The damage of electronic components during manufacturing. Photocopier and laser printer operation.
Complete answer:
Electrostatics: Electrostatics branch of physics deals with the electric effect of static charge is called electrostatics. The effect of stationary or slow- moving electric charges. Electrostatic cause from the force that electric charge exerts on each other and are described by Coulomb’s law.
Coulomb’s Law: It begins with the magnitude of the electrostatic force between two-point charges $q\,\,and\,Q$ . It is convenient to label one of these charges, $q$ as a test charge and $Q$ a source charge. As we develop the theory, more source charges will be beaded. If $r$ is the distance between two charges, then the force of electrostatic formula is:
$
F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{qQ}}{{{r^2}}} = {k_e}\dfrac{{qQ}}{{{r^2}}} \\
F = k\dfrac{{{q_1}{q_2}}}{{{d^2}}} \\
$
Force between $2$ charges $0.06m$ apart is $5N$
We know, $F = \dfrac{{K{q_1}{q_2}}}{{{r^2}}}$
$S = \dfrac{{K{q_1}{q_2}}}{{{{\left( {0.06} \right)}^2}}}$
$K{q_1}{q_2} = 0.018..................\left( 1 \right)$
${F^1} = \dfrac{{K{q_1}{q_2}}}{{{{\left[ {0.06 - 2\left( {0.01} \right)} \right]}^2}}}$ charge is moved with regard to the other by $0.04m$
${F^1} = \dfrac{{0.018}}{{{{\left( {0.04} \right)}^2}}}$
${F^1} = 11.25N$
Force will become $11.25$ newtons.
Hence, the correct option is B $11.25N$
Note: Coulomb’s law of physics is the incident that measures the amount of force between two stationaries, electrically charged particles. The electric force between charge bodies at rest is normally called electrostatic force. There are many examples of electrostatic events, the attraction of the plastic wrapped on hand after removing it from a package, the attraction of paper to a charged scale, the apparently unconstrained explosion of grain silos. The damage of electronic components during manufacturing. Photocopier and laser printer operation.
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