Numerical Problems: Electric Charges and Fields | Physics Class 12 - NEET PDF Download

Q1. An electric dipole of length 2 cm is placed with its axis making an angle of 30° to a uniform electric field 10⁵ N/C. If it experiences a torque of 10√3 Nm, then potential energy of dipole will
Sol: τ = P E sin 30
10√3 = PE / 2
PE = 20√3
Potential Energy = –PE cos 30
Potential energy = –20√3 × √3 / 2
= –10 × 3 = –30 J

Q2. Calculate the net force acting on the charge present at the origin.

Sol: F'' = √2 F
F'' = k q q₁ / a²
F' = k q q₁ / (√2 a)² = k q q₁ / 2a²
∴ F_Net = F'' – F'
= k q q₁ / a² (√2 – 1/2)

Q3. A hollow charged metal sphere has radius r. If the potential difference between its surface and a point at a distance 3r from the centre is V, then electric field intensity at distance 3r from the centre is
Sol: V_A – V_B = kQ / r – kQ / 3r
= kQ / 3r, V = 2 kQ / 3r
E = kQ / (3r)² = kQ / 9r² = 3 r V / 2 9r². E = V / 6r

Q4. If two like charges of magnitude 1 × 10^-9 coulomb and 9 × 10^-9 coulomb are separated by a distance of 1 meter, then the point on the line joining the charges, where the force experienced by a charge placed at that point is zero, is
Sol: F₂₁ = F₂₃
or k q q₁ / x² = k q q₂ / (1 - x)²
or 1 × 10^-9 / x² = 9 × 10^-9 / (1 - x)²
or 1 / x² = 9 / (1 - x)²
or 1 / x = 3 / (1 - x)
or 1 - x = 3x
or x = 0.25 m from 1 × 10^-9C
∴ From 9 × 10^-9C, distance
= 1 - x
= 1 - 0.25
= 0.75

Q5. A charge Q has to be divided between two solid spheres of radius 'R' which are at distance d from each other (d >> R). What should be the value of charge, which we should place on spheres, so that the force of attraction between them is maximum?
Sol: Let us place a quantum of charge 'q' on the first sphere, so we have charge Q-q on the other sphere. The force of attraction between the spheres is
F = (Q - q)q / 4πϵ₀d²
dF/dq = 1 / 4πϵ₀d² [(Q - q) - q] = 0
∴ q = Q/2

Q6. The electric potential V at any point (x, y, z) in space is given by V = 6x² volt, where all the distances are measured in metre. The electric field at the point (1 m, 0, 2 m) is
Sol: The potential V is a scalar function, whereas the field E is a vector function. The three components of E are given as
E_x = - ∂V / ∂x = - ∂ / ∂x (6x²) = -12x
E_y = - ∂V / ∂y = - ∂ / ∂y (6x²) = 0
E_z = - ∂V / ∂z = - ∂ / ∂z (6x²) = 0
∴ E = -12xî + 0ĵ + 0k̂ = -12xî
at the given point, x = 1,
∴ E = -12î

Q7. How many electrons should be removed from a coin of mass 1.6 gm, so that it may float in electric field intensity 10⁵ N/C directed upwards?

Sol: qE = mg
neE = mg
n = mg / eE = 1.6×10^-3 × 9.8 / 1.6×10^-19 × 10⁵
n = 9.8 × 10⁷

Q8. An electric dipole, made up of a positive and a negative charge, each of 1μC and placed at a distance 2 cm apart, is placed in an electric field 10⁵ N/C. Compute the maximum torque which the field can exert on the dipole, and the work that must be done to turn the dipole from a position θ = 0° to θ = 180°.
Sol: The torque exerted by an electric field E on a dipole of moment p is given by
τ = pE sinθ,
where θ is the angle which the dipole is making with the field.
τ is a maximum, when θ = 90°. That is
∴ τ_max = pE
Here p = q(2ℓ) = 1×10^-6 × 0.02C/m and E = 10⁵N/C
∴ τ_max = 1×10^-6 × 0.02 × 10⁵ = 2×10^-3N-m
The work done in rotating the dipole from an angle θ₀ to θ is given by
W = ∫_θ₀^θ pE sinθ dθ = pE (cosθ₀ - cosθ)
Here θ₀ = 0° and θ = 180°
∴ W = pE (cos0° - cos180°) = 2pE = 4×10^-3N-m or Joule

Q9. Potential in the x-y plane is given as V = 5(x² + xy) volts. The electric field at the point (1, -2) will be
Sol: E_x = - ∂V / ∂x = - (10x + 5y) = -10 + 10 = 0
E_y = - ∂V / ∂y = -5x = -5
∴ E = -5ĵ V/m.

Q10. A long string with a charge of λ per unit length passes through an imaginary cube of edge a. The maximum flux of the electric field through the cube will be
Sol: The maximum length of the string which can fit into the cube is √3a, equal to its body diagonal. The total charge inside the cube is √3λa, and hence the total flux through the cube is √3λa / ε₀.