The capacitance of a spherical conductor of radius R is:
Answer: B
Capacitance of isolated sphere: C = 4πε₀R, where R is radius. Since k = 1/(4πε₀), C = R/k
Q.102Medium
The potential energy of electric dipole in uniform field E is:
Answer: A
Potential energy of dipole: U = -p·E = -pE cosθ, where θ is angle between p and E. Minimum at θ = 0
Q.103Medium
Two point charges +8μC and -2μC are separated by 3 m. The electric potential is zero at a point on the line joining them, located at distance x from the +8μC charge. Find x:
Answer: C
For zero potential: 8/(x) = 2/(3-x). Solving: 8(3-x) = 2x → 24 = 10x → x = 2.4 m
Q.104Medium
A parallel plate capacitor with plate separation d and area A is charged to voltage V. A dielectric of dielectric constant K is inserted between the plates. The change in stored energy is:
Answer: A
If isolated (constant charge): U ∝ 1/C, and C increases by K, so U decreases by K. If connected to battery (constant V): U ∝ C, increases by K. Given 'charged' implies isolated.
Q.105Medium
Two conducting spheres of radii r₁ and r₂ (r₁ > r₂) have charges Q₁ and Q₂. They are connected by a wire. The final charge distribution will be such that:
Answer: D
Connected spheres have same potential. V = kQ/r, so Q ∝ r. Both statements are equivalent and correct.
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Q.106Medium
The electric field on the axis of a uniformly charged ring of radius a and total charge Q at distance x from the center is:
Answer: A
By symmetry, radial components cancel. Axial component: E = kQx/(a² + x²)^(23)
Q.107Medium
Two identical metal spheres with charges +Q and -3Q are brought into contact and then separated by distance r. The electrostatic force between them is:
Answer: A
After contact: charge on each = (Q - 3Q)/2 = -Q. Force F = k(-Q)(-Q)/r² = kQ²/r², attractive (opposite signs).
Q.108Medium
The electric field intensity at distance r from a line charge with linear charge density λ is given by:
Answer: B
Using Gauss's law with cylindrical surface: E × 2πr = λL/ε₀, hence E = λ/(2πε₀r)
Q.109Medium
A parallel plate capacitor has capacitance C and is charged to voltage V. If voltage is doubled and a dielectric K is inserted, the energy stored becomes:
Answer: A
U = ½C'V'² where C' = KC and V' = 2V. So U = ½(KC)(2V)² = 4CKV²
Q.110Medium
A metallic sphere of radius R is charged such that surface charge density is σ. If the sphere is surrounded by a dielectric medium of dielectric constant K, how does the electric field just outside the surface change?
Answer: B
The electric field just outside a conductor in a dielectric medium is E = σ/(Kε₀), so it decreases by the factor of dielectric constant K compared to vacuum.
Q.111Medium
Two fixed point charges Q₁ = +2μC and Q₂ = -2μC are separated by 2 m. A test charge +1μC is moved along the perpendicular bisector. At what distance from the midpoint (on the bisector) is the electric field maximum?
Answer: B
For an electric dipole, the field along the perpendicular bisector is maximum at distance d = a/√2, where 2a is the separation between charges. Here, a = 1 m, so maximum field is at 1/√2 m.
Q.112Medium
A parallel plate capacitor is filled with two dielectrics of thickness d/2 each and dielectric constants K₁ and K₂. What is the equivalent capacitance if the original capacitance (empty) was C₀?
Answer: D
Dielectrics in series combine like capacitors in series. Total capacitance = (ε₀A × 2K₁K₂)/(d(K₁ + K₂)) = C₀ × 2K₁K₂/(K₁ + K₂).
Q.113Medium
A hollow conducting sphere of radius R carries charge Q. A point charge q is placed at the center. What is the electric field at distance r from center (r > R)?
Answer: B
By Gauss's law, the field outside depends on the total enclosed charge Q + q. The field is E = k(Q + q)/r² for r > R.
Q.114Medium
The electric field just outside a conductor surface is perpendicular to the surface. This is because:
Answer: B
If there were a tangential component of electric field at the surface, charges would move tangentially, violating electrostatic equilibrium. Thus, only the normal component exists.
Q.115Medium
Three capacitors of capacitance C each are connected in a combination where two are in parallel and this combination is in series with the third. The equivalent capacitance is:
Answer: B
Two capacitors C in parallel: C_parallel = 2C. This in series with C: C_eq = (2C × C)/(2C + C) = 2C/3.
Q.116Medium
A point charge is brought from infinity to a point in an electric field. Work done depends on:
Answer: C
Electric force is conservative; work depends only on initial and final positions (potentials), not on the path taken. W = q(V_initial - V_final).
Q.117Medium
A uniformly charged ring of radius R carries total charge Q. What is the electric field at a point on the axis at distance x from the center?
Answer: A
Using Coulomb's law and symmetry, the radial components cancel. The axial component gives E = kQx/(x² + R²)^(23).
Q.118Medium
An electric dipole of moment p is placed in a uniform electric field E at angle θ to the field. The potential energy is:
Answer: A
Potential energy of dipole in electric field: U = -p·E = -pE cos θ. Minimum at θ = 0 (aligned).
Q.119Medium
A non-conducting sphere of radius R is uniformly charged with charge density ρ. What is the electric field at distance r from center (r < R)?
Answer: A
Using Gauss's law with spherical symmetry: E(4πr²) = (ρ × 4πr³/3)/ε₀, giving E = ρr/(3ε₀).
Q.120Medium
The work done to move a charge q from point A (potential V_A) to point B (potential V_B) is:
Answer: B
Work done by external agent = q(V_B - V_A). If V_B > V_A, positive work is needed.
