StudentAidUnit 15: Electrostatics — Textbook Exercise Numericals
10th Class Physics · Unit 15: Electrostatics
15.1.A positive test charge of 25 μC is placed in an electric field. The force on it is 0.500 N. What is the magnitude of the electric field at the location of the test charge?
Given
Test charge
q = 25 μC = 25 × 10-6 C
Force on the charge
F = 0.500 N
To find
Magnitude of the electric field
E = ?
Formula
The formula for electric field intensity is:
E = Fq0
Working
Substituting the values given, we have:
E = 0.50025 × 10-6
E = 20000 N C-1
E = 2 × 104 N C-1
Result
The magnitude of the electric field is:
E = 2.0 × 104 N C-1
15.2.Two point charges, q₁ = 8 μC and q₂ = 4 μC, are placed at a distance of 120 cm. What will be the Coulomb's force between them? Also, find the nature of the force.
Given
Charge 1
q1 = 8 μC = 8 × 10-6 C
Charge 2
q2 = 4 μC = 4 × 10-6 C
Distance between charges
r = 120 cm = 1.20 m
Coulomb's constant
k = 9 × 109 N m2 C-2
To find
Coulomb's force
F = ?
Nature of the force
Formula
Using Coulomb's Law:
F = kq1 q2r2
Working
Substituting the values:
F = (9 × 109)(8 × 10-6) × (4 × 10-6)(1.20)2
F = (9 × 109)32 × 10-121.44
F = (9 × 109) × (22.22 × 10-12)
F = 199.98 × 10-3 N
F ≈ 0.2 N
Since both charges are positive, the force is repulsive.
Result
The Coulomb's force is 0.2 N and it is repulsive.
F ≈ 0.2 N (repulsive)
15.3.Two identical charges repel each other with a force of 0.2 N when they are 9 cm apart. Find the force between the same charges when they are 3 cm apart.
Given
Initial force
F1 = 0.2 N
Initial distance
r1 = 9 cm = 0.09 m
Final distance
r2 = 3 cm = 0.03 m
Charges q₁ and q₂ remain the same.
To find
Final force
F2 = ?
Formula
From Coulomb's law:
F ∝ 1r2
Therefore:
F2F1 = r12r22
F2 = F1 × r12r22
Working
Substituting the values:
F2 = 0.2 × (0.09)2(0.03)2
F2 = 0.2 × 0.00810.0009
F2 = 1.8 N
Result
The force when they are 3 cm apart is 1.8 N.
F2 = 1.8 N
15.4.A point charge q = 10 μC is placed in air. Calculate the electric field at a distance of 0.5 metres from the charge.
Given
Point charge
q = 10 × 10-6 C
Distance from charge
r = 0.5 m
Coulomb's constant
k = 9 × 109 N m2 C-2
To find
Electric field intensity
E = ?
Formula
The formula for electric field due to a point charge is:
E = kqr2
Working
Substituting the values:
E = (9 × 109)10 × 10-6(0.5)2
E = (9 × 109)10 × 10-60.25
E = (9 × 109) × (40 × 10-6)
E = 360 × 103 N C-1
E = 3.6 × 105 N C-1
Result
The electric field at 0.5 m is:
E = 3.6 × 105 N C-1
15.5.Two small equally charged metal spheres having charge 3 μC are brought close together such that the distance between them is 2.0 cm. Calculate the magnitude of repulsive force that each sphere exerts on the other.
Given
Charge on each sphere
q = 3 μC = 3 × 10-6 C
Distance between spheres
r = 2.0 cm = 0.02 m
Coulomb's constant
k = 9 × 109 N m2 C-2
To find
Repulsive force
F = ?
Formula
Using Coulomb's Law:
F = kq × qr2
Working
Substituting the values, we have:
F = (9 × 109)(3 × 10-6)2(0.02)2
F = (9 × 109)9 × 10-120.0004
F = (9 × 109) × (2.25 × 10-8)
F = 20.25 × 101 N
F = 202.5 N
Result
The magnitude of the repulsive force is 202.5 N.
F = 202.5 N
15.6.Two small equally charged metal spheres repel each other with a force of 0.5 N when placed 3 cm apart in the air. Calculate the charge on each sphere.
15.7.A force of 85 N acts between two charges placed 3.2 cm apart. If one charge is 25 μC, determine the other charge.