
How does the electric force between two charged objects change?
Answer
545.7k+ views
Hint: The attractive and repulsive forces between particles which are caused due to their electric charges is known as the electrostatic force. Conventionally the electric force between stationary charged bodies is known as the electrostatic force. It is also referred to as Coulomb's force.
Complete step by step answer:
As we know the electrostatic force between two charged particles is calculated as Coulomb's force as is calculated as
$F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}}$ or $F = {k_e}\dfrac{{{q_1}{q_2}}}{{{r^2}}}$
Where F is the electrostatic force
${k_e}$is Coulomb's constant and its value is $8.8987 \times {10^9}N{m^2}{C^{ - 2}}$
${q_1}{q_2}$are the charges on the particles
And r is the distance between the particles.
Now here we can see that the electrostatic force (F) depends upon the magnitude of the charges on two particles and inversely depends upon the distance between two charges.
Here it is evident that $F \propto \dfrac{1}{{{r^2}}}$
So, we can conclude here that the electrostatic force (F) is inversely proportional to the square of the distance between two particles. As the distance between two-particles decreases, force between them will increase.
Hence, the electrostatic force (F) depends upon the magnitude of the charges on two particles and inversely proportional to the square of the distance between two charges.
Note:
• The magnitude of the charge should be kept along with its sign.
• The electrostatic charge may be attractive and repulsive depending upon the nature of the charge on the two-body.
• Like charges repel and unlike charges attract each other.
Complete step by step answer:
As we know the electrostatic force between two charged particles is calculated as Coulomb's force as is calculated as
$F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}}$ or $F = {k_e}\dfrac{{{q_1}{q_2}}}{{{r^2}}}$
Where F is the electrostatic force
${k_e}$is Coulomb's constant and its value is $8.8987 \times {10^9}N{m^2}{C^{ - 2}}$
${q_1}{q_2}$are the charges on the particles
And r is the distance between the particles.
Now here we can see that the electrostatic force (F) depends upon the magnitude of the charges on two particles and inversely depends upon the distance between two charges.
Here it is evident that $F \propto \dfrac{1}{{{r^2}}}$
So, we can conclude here that the electrostatic force (F) is inversely proportional to the square of the distance between two particles. As the distance between two-particles decreases, force between them will increase.
Hence, the electrostatic force (F) depends upon the magnitude of the charges on two particles and inversely proportional to the square of the distance between two charges.
Note:
• The magnitude of the charge should be kept along with its sign.
• The electrostatic charge may be attractive and repulsive depending upon the nature of the charge on the two-body.
• Like charges repel and unlike charges attract each other.
Recently Updated Pages
Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

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

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

What is 1s 2s 2p 3s 3p class 11 chemistry CBSE

An example of chemosynthetic bacteria is A E coli B class 11 biology CBSE

10 examples of friction in our daily life

