two insulating charge copper sphere A and B have their centred
Answers
Answer:
- Force = 0.1521 N
Given Information:
- Distance of Separation = 50 cm = 0.5 m
- Charge on each sphere = 6.5 × 10⁻⁷ C
To find:
- Force of Electrostatic Repulsion
Steps:
where,
'q1' and 'q2' are the charges on each sphere, 'r' is the separation distance and 'k' is Coulomb constant which has a value 9 × 10⁹ units.
Substituting the values we get:
Hence the force of electrostatic repulsion is 0.1521 N.
Required answer -
Question -
⚕ Two insulating charge copper sphere A and B have their centred separated by 50 cm. It is the mautual electrostatic repulsion if the on each is 6.5 × 10-⁷C. Find force.
Given that -
⚕ Separation distance = 50 cm or 0.5 m
⚕ Charges on each copper sphere = 6.5 × 10-⁷C
To find -
⚕ Force of the electrostatic repulsion
Solution -
⚕ Force of the electrostatic repulsion = 0.1521N
Using concept -
⚕ Formula to find electrostatic repulsion
Using formula -
Where,
⚕ k denotes colounb constant whose value is 9 × 10⁹ units.
⚕ q₁ and q₂ denotes the charges of the spheres.
⚕ r denotes distance of separation.
Now let's put the values according to the formula,
More knowledge -
⚔ Physics ⚔
Law of Exponents -