Three identical charges q are placed at the vertices of an equilateral triangle of side a. The electric potential at the centroid is:
Answer: A
Distance from each vertex to centroid = a/√3. Total potential = 3 × kq/(a/√3) = 3√3kq/a. Correction: V = 3kq√3/a ≈ 3kq/a for approximation
Q.2Medium
A uniformly charged infinite plane sheet has surface charge density σ. The electric field due to the sheet is:
Answer: A
Using Gauss's law for an infinite plane sheet: E = σ/(2ε₀). The field is independent of distance and uniform on both sides
Q.3Medium
Two concentric spheres have radii r₁ (inner) and r₂ (outer). The inner sphere has charge +Q and outer sphere has charge -Q. The electric field in the region r₁ < r < r₂ is:
Answer: B
By Gauss's law, in the region between spheres, only inner charge +Q contributes. E = kQ/r², directed radially outward
Q.4Medium
A parallel plate capacitor is charged to voltage V and then isolated. If the plate separation is doubled, the energy stored becomes:
Answer: C
For isolated capacitor, charge Q remains constant. U = Q²/(2C) = Q²d/(2ε₀A). Energy is proportional to d, so doubling d doubles energy... correction: U = CV²/2, but Q is constant so U = Q²/(2C) ∝ d, energy increases
Q.5Medium
A charged particle moves from point A to point B in electrostatic field. The work done by electrostatic force is independent of path because:
Answer: A
Electrostatic force is conservative, meaning work depends only on initial and final positions, not the path taken
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Q.6Medium
A conducting sphere of radius R₁ inside a conducting spherical shell of radius R₂ (R₁ < R₂) has charge Q. The electric field for r > R₂ is:
Answer: B
By Gauss's law, field outside depends only on total enclosed charge Q. E = kQ/r² for r > R₂
Q.7Medium
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.8Medium
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.9Medium
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.10Medium
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.11Medium
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.
Q.12Medium
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.13Medium
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.14Medium
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.15Medium
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.16Medium
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.17Medium
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.18Medium
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.19Medium
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.20Medium
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.
