Imagine a figure showing a square of side 10 cm with four charges placed at its corners. O is the centre of the square. Where, Four point charges qA= 2 µC, qB = -5 µC, qC = 2 µC, and qD = -5 µC are located at the corners of a square ABCD of side 10 cm. (Sides) AB = BC = CD = AD = 10 cm (Diagonals) ARead more
Imagine a figure showing a square of side 10 cm with four charges placed at its corners. O is the centre of the square.
Where, Four point charges qA= 2 µC, qB = -5 µC, qC = 2 µC, and qD = -5 µC are located at the corners of a square ABCD of side 10 cm.
(Sides) AB = BC = CD = AD = 10 cm
(Diagonals) AC = BD = 10√ 2 cm
AO = OC = DO = OB = 5√ 2 cm A charge of amount lµC is placed at point O.
Force of repulsion between charges placed at corner A and centre O is equal in magnitude but opposite in direction relative to the force of repulsion between the charges placed at corner C and centre O. Hence, they will cancel each other. Similarly, force of attraction between charges placed at corner B and centre O is equal in magnitude but opposite in direction relative to the force of attraction between the charges placed at corner D and centre O. Hence, they will also cancel each other. Therefore, net force caused by the four charges placed at the corner of the square on 1 µC charge at centre O is zero.
Given the electrostatic force on a small sphere of charge 0.4 µC due to another small sphere of charge - 0.8 µC in air is 0.2 N. (a) What is the distance between the two spheres? (b) What is the force on the second sphere due to the first? (a) Electrostatic force on the first sphere, F = 0.2 N ChargRead more
Given the electrostatic force on a small sphere of charge 0.4 µC due to another small sphere of charge – 0.8 µC in air is 0.2 N.
(a) What is the distance between the two spheres?
(b) What is the force on the second sphere due to the first?
(a) Electrostatic force on the first sphere, F = 0.2 N Charge on this sphere, q1 = 0.4 µC = 0.4 X 10-6 C
Charge on the second sphere,
q2 = – 0.8 µC = – 0.8 x 10-6 C
between the spheres is given by the relation
Since Electrostatic force between two charges q1 and q2 separated by a distance of r is given by expression:
F=1 /4πε0 x q1q2/r²
Where ε0 =Permittivity of free space and value of
1 /4πε0 =9 X 109 Nm 2C⁻²
Therefore,
r2= 1 /4πε0 x (q1q2)/F
r2= 9 x 109 x 0.4 x 10-6 x 8 x 10-6 /0.2
r2= 144 x 10-4
r=√(144 x 10-4)=(12×10-2)=0.12
The distance between the two spheres is 0.12 m.
(b) Both the spheres attract each other with the same force. Therefore, the force on the second sphere due to the first is 0.2 N.
Since force between two charges q1 and q2 separated by a distance of r is given by expression: F=1 /4πε0 x q1q2/r² Where ε0 =Permittivity of free space and value of 1 /4πε0 =9 X 109 Nm 2C⁻² and given Charge on the first sphere, q1= 2 x 10-7 C Charge on the second sphere, q2= 3 x 10-7 C DistancRead more
Since force between two charges q1 and q2 separated by a distance of r is given by expression:
F=1 /4πε0 x q1q2/r²
Where ε0 =Permittivity of free space and value of
1 /4πε0 =9 X 109 Nm 2C⁻²
and given
Charge on the first sphere, q1= 2 x 10-7 C Charge on the second sphere,
q2= 3 x 10-7 C Distance between the spheres, r = 30 cm = 0.3 m
Therefore
F=9 x 109 x (2 x 10-7 x3 x 10-7 )/(0.3 )²
F=6 X 10⁻ ³ N
Hence, force between the two small charged spheres is 6 x 10-3 N. The charges are of same nature. Hence, force between them will be repulsive.
Four point charges q₁= 2 μC, q₂ = –5 μC, q₃ = 2 μC, and q₄ = –5 μC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 μC placed at the centre of the square?
Imagine a figure showing a square of side 10 cm with four charges placed at its corners. O is the centre of the square. Where, Four point charges qA= 2 µC, qB = -5 µC, qC = 2 µC, and qD = -5 µC are located at the corners of a square ABCD of side 10 cm. (Sides) AB = BC = CD = AD = 10 cm (Diagonals) ARead more
Imagine a figure showing a square of side 10 cm with four charges placed at its corners. O is the centre of the square.
Where, Four point charges qA= 2 µC, qB = -5 µC, qC = 2 µC, and qD = -5 µC are located at the corners of a square ABCD of side 10 cm.
(Sides) AB = BC = CD = AD = 10 cm
(Diagonals) AC = BD = 10√ 2 cm
AO = OC = DO = OB = 5√ 2 cm A charge of amount lµC is placed at point O.
Force of repulsion between charges placed at corner A and centre O is equal in magnitude but opposite in direction relative to the force of repulsion between the charges placed at corner C and centre O. Hence, they will cancel each other. Similarly, force of attraction between charges placed at corner B and centre O is equal in magnitude but opposite in direction relative to the force of attraction between the charges placed at corner D and centre O. Hence, they will also cancel each other. Therefore, net force caused by the four charges placed at the corner of the square on 1 µC charge at centre O is zero.
See lessThe electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge –0.8 μC in air is 0.2 N. (a) what is the distance between the two spheres? (b) What is the force on the second sphere due to the first?
Given the electrostatic force on a small sphere of charge 0.4 µC due to another small sphere of charge - 0.8 µC in air is 0.2 N. (a) What is the distance between the two spheres? (b) What is the force on the second sphere due to the first? (a) Electrostatic force on the first sphere, F = 0.2 N ChargRead more
Given the electrostatic force on a small sphere of charge 0.4 µC due to another small sphere of charge – 0.8 µC in air is 0.2 N.
(a) What is the distance between the two spheres?
(b) What is the force on the second sphere due to the first?
(a) Electrostatic force on the first sphere, F = 0.2 N Charge on this sphere, q1 = 0.4 µC = 0.4 X 10-6 C
Charge on the second sphere,
q2 = – 0.8 µC = – 0.8 x 10-6 C
between the spheres is given by the relation
Since Electrostatic force between two charges q1 and q2 separated by a distance of r is given by expression:
F=1 /4πε0 x q1q2/r²
Where ε0 =Permittivity of free space and value of
1 /4πε0 =9 X 109 Nm 2C⁻²
Therefore,
r2= 1 /4πε0 x (q1q2)/F
r2= 9 x 109 x 0.4 x 10-6 x 8 x 10-6 /0.2
r2= 144 x 10-4
r=√(144 x 10-4)=(12×10-2)=0.12
The distance between the two spheres is 0.12 m.
(b) Both the spheres attract each other with the same force. Therefore, the force on the second sphere due to the first is 0.2 N.
See lessWhat is the force between two small charged spheres having charges of 2 × 10⁻⁷ C and 3 × 10 ⁻⁷ C placed 30 cm apart in air?
Since force between two charges q1 and q2 separated by a distance of r is given by expression: F=1 /4πε0 x q1q2/r² Where ε0 =Permittivity of free space and value of 1 /4πε0 =9 X 109 Nm 2C⁻² and given Charge on the first sphere, q1= 2 x 10-7 C Charge on the second sphere, q2= 3 x 10-7 C DistancRead more
Since force between two charges q1 and q2 separated by a distance of r is given by expression:
F=1 /4πε0 x q1q2/r²
Where ε0 =Permittivity of free space and value of
1 /4πε0 =9 X 109 Nm 2C⁻²
and given
Charge on the first sphere, q1= 2 x 10-7 C Charge on the second sphere,
q2= 3 x 10-7 C Distance between the spheres, r = 30 cm = 0.3 m
Therefore
F=9 x 109 x (2 x 10-7 x3 x 10-7 )/(0.3 )²
F=6 X 10⁻ ³ N
Hence, force between the two small charged spheres is 6 x 10-3 N. The charges are of same nature. Hence, force between them will be repulsive.