Question 1.12: (a) Two insulated charged copper spheres A and B have their centers separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 × 10 −7 C? The radii of A and B are negligible compared to the distance of separation. (b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?
Class 12 - Physics - Electric Charges And Fields Electric Charges And Fields Page-46
Answers
Answered by
9
Final Answer :a) 1.5 * 10^-2N
b) 24 * 10^-2N
Understanding :
1) Both spheres will repel each other as they have same charges.
2) It is mentioned that distance between them is much larger than their radius
=> We can neglect any other kind of interaction from the surface charges of sphere between.
Hence, We can assume both as point charges.
Steps :
a) Charge, q = 6.5 * 10^-7 C
Distance, r = 50cm = 0.5m
By Coulomb law of point charges,
F = kqq/r^2
=
Newtons.
Now, we see that no. of significant figures in charge is 2.
So, on solving Force also must have max no. of significant figures as 2 with round off.
Therefore,
F = 1.5 * 10^(-2) N
(b) We know that,
Here, Force is directly proportional to product of two charges.
So, when charge on each sphere is doubles, then Force becomes 2* 2 = 4 times.
And, When
Force is inversely proportional to square of distance between them,
So, when distance between them is halved, then Force becomes 1/(1/2)^2 = 4 times.
Overall ,
Required force becomes 4 * 4 = 16 times Original.
That is, F (req.) = 16 * 1.5 * 10^-2.
= 24 * 10^-2 N
b) 24 * 10^-2N
Understanding :
1) Both spheres will repel each other as they have same charges.
2) It is mentioned that distance between them is much larger than their radius
=> We can neglect any other kind of interaction from the surface charges of sphere between.
Hence, We can assume both as point charges.
Steps :
a) Charge, q = 6.5 * 10^-7 C
Distance, r = 50cm = 0.5m
By Coulomb law of point charges,
F = kqq/r^2
=
Newtons.
Now, we see that no. of significant figures in charge is 2.
So, on solving Force also must have max no. of significant figures as 2 with round off.
Therefore,
F = 1.5 * 10^(-2) N
(b) We know that,
Here, Force is directly proportional to product of two charges.
So, when charge on each sphere is doubles, then Force becomes 2* 2 = 4 times.
And, When
Force is inversely proportional to square of distance between them,
So, when distance between them is halved, then Force becomes 1/(1/2)^2 = 4 times.
Overall ,
Required force becomes 4 * 4 = 16 times Original.
That is, F (req.) = 16 * 1.5 * 10^-2.
= 24 * 10^-2 N
Answered by
4
(a) charge on each copper sphere , q = 6.5 × 10⁻⁷C
speration between copper spheres , r = 50 cm
According to Coulombs law,
F = Kq₁q₂/r²
Here, q₁ = q₂ = q = 6.5 × 10⁻⁷C
r = 50cm = 0.5 m
∴ F = 9 × 10⁹ × 6.5 × 10⁻⁷ × 6.5 × 10⁻⁷/(0.5)²
= 9 × 6.5 × 6.5 × 10⁻⁵/0.25
= 9 × 13 × 13 × 10⁻⁵
= 1521 × 10⁻⁵
= 1.521 × 10⁻² N
Hence, mutual force of Electrostatic repulsion = 1.521 × 10⁻² N
(b) charge on each sphere is doubled
e.g., charge on each sphere , q' = 2q = 2 × 6.5 × 10⁻⁷ = 13 × 10⁻⁷ C
speration between them is halved
e.g., r' = r/2 =50cm/2 = 25cm = 0.25 m
Now, Force = kq'²/r'²
= 9 × 10⁹ × (13 × 10⁻⁷)²/(0.25)²
= 9 × 169 × 10⁻⁵/0.25 × 0.25
= 16 × 9 × 169 × 10⁻⁵
= 16 × 1521 × 10⁻⁵
≈ 0.24N
Hence, force will be 0.24N
speration between copper spheres , r = 50 cm
According to Coulombs law,
F = Kq₁q₂/r²
Here, q₁ = q₂ = q = 6.5 × 10⁻⁷C
r = 50cm = 0.5 m
∴ F = 9 × 10⁹ × 6.5 × 10⁻⁷ × 6.5 × 10⁻⁷/(0.5)²
= 9 × 6.5 × 6.5 × 10⁻⁵/0.25
= 9 × 13 × 13 × 10⁻⁵
= 1521 × 10⁻⁵
= 1.521 × 10⁻² N
Hence, mutual force of Electrostatic repulsion = 1.521 × 10⁻² N
(b) charge on each sphere is doubled
e.g., charge on each sphere , q' = 2q = 2 × 6.5 × 10⁻⁷ = 13 × 10⁻⁷ C
speration between them is halved
e.g., r' = r/2 =50cm/2 = 25cm = 0.25 m
Now, Force = kq'²/r'²
= 9 × 10⁹ × (13 × 10⁻⁷)²/(0.25)²
= 9 × 169 × 10⁻⁵/0.25 × 0.25
= 16 × 9 × 169 × 10⁻⁵
= 16 × 1521 × 10⁻⁵
≈ 0.24N
Hence, force will be 0.24N
Similar questions