physics : class12 Integration of Elementary Electric Fields : Superposition Principle : Electrostatics MCQ Questions and Answers
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Integration of Elementary Electric Fields
MCQ questions and answers.
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MCQ Questions
1. A uniformly charged rod of length L lies along the x-axis from x = 0 to x = L, with linear charge density λ. What is the electric field at the point x = −a, where a > 0?
A.
kλ(1/a − 1/(a + L)) directed along the negative x-axis
B.
kλ(1/a + 1/(a + L)) directed along the negative x-axis
C.
kλ ln((a + L)/a) directed along the negative x-axis
D.
kλL/[a(a + L)] directed along the positive x-axis
Correct Answer:
kλ(1/a − 1/(a + L)) directed along the negative x-axis
Explanation:
For an element dq = λ dx at position x, dE = kλ dx/(a + x)^2. All elements produce fields in the negative x-direction, so E = kλ∫₀ᴸ dx/(a + x)^2 = kλ[1/a − 1/(a + L)].
2. A uniformly charged semicircular arc of radius R lies in the upper half-plane and has linear charge density λ > 0. What is the electric field at its center?
A.
2kλ/R in the positive y-direction
B.
2kλ/R in the negative y-direction
C.
πkλ/R in the negative y-direction
D.
kλ/R in the positive x-direction
Correct Answer:
2kλ/R in the negative y-direction
Explanation:
For an element dq = λR dθ, dE = kλ dθ/R. The horizontal components cancel by symmetry, while the vertical component is dE_y = −(kλ/R)sinθ dθ. Integrating from 0 to π gives E_y = −(kλ/R)∫₀π sinθ dθ = −2kλ/R.
3. A uniformly charged thin disk of radius R has surface charge density σ. At a point on its axis a distance z > 0 from its center, which expression is obtained by integrating the contributions of concentric rings?
A.
σ/(2ε₀) [1 + z/√(z² + R²)]
B.
σ/(2ε₀) [1 − z/√(z² + R²)]
C.
σR/(2ε₀√(z² + R²))
D.
σ/(2ε₀) [z/√(z² + R²) − 1]
Correct Answer:
σ/(2ε₀) [1 − z/√(z² + R²)]
Explanation:
A ring of radius r contributes dE_z = (1/(4πε₀))(σ2πr dr)z/(z²+r²)^(3/2). Integrating r from 0 to R gives E_z = σz/(2ε₀)∫₀ᴿ r dr/(z²+r²)^(3/2) = σ/(2ε₀)[1 − z/√(z²+R²)].
4. A rod extends from x = 0 to x = L along the x-axis and has nonuniform linear charge density λ(x) = λ₀x/L. What is the electric field at x = −a, where a > 0?
A.
kλ₀/L [ln((a + L)/a) + a/(a + L) − 1] in the negative x-direction
B.
kλ₀/L [ln((a + L)/a) − a/(a + L) + 1] in the negative x-direction
C.
kλ₀/L [1/a − 1/(a + L)] in the negative x-direction
D.
kλ₀/L [ln((a + L)/a) + 1 − a/(a + L)] in the positive x-direction
Correct Answer:
kλ₀/L [ln((a + L)/a) + a/(a + L) − 1] in the negative x-direction
Explanation:
Here dq = (λ₀x/L)dx and each element produces dE = kλ₀x dx/[L(a+x)^2] toward negative x. Using x/(a+x)^2 = 1/(a+x) − a/(a+x)^2, integration from 0 to L gives the stated result.
5. A uniformly charged ring of radius R and total charge Q lies in the xy-plane. What is the electric field at a point on its axis a distance z from the center?
A.
kQ/(R² + z²) directed along the axis
B.
kQz/(R² + z²)^(3/2) directed along the axis
C.
kQ/(R² + z²)^(3/2) directed radially outward
D.
kQz/(R² + z²) directed radially outward
Correct Answer:
kQz/(R² + z²)^(3/2) directed along the axis
Explanation:
Every element is at the same distance √(R²+z²) from the point. Radial components cancel around the ring, while the axial component is dE_z = k z dq/(R²+z²)^(3/2). Integration over the ring replaces ∫dq by Q.
6. A uniformly charged finite rod of length L is centered on the x-axis, with linear charge density λ. What is the electric field at a point on its perpendicular bisector a distance y from the center?
A.
kλL/[y√(y² + L²/4)] in the direction away from the rod
B.
2kλL/[y√(y² + L²/4)] in the direction away from the rod
C.
2kλL/[√(y² + L²/4)] in the direction toward the rod
D.
kλL/[y² + L²/4] in the direction away from the rod
Correct Answer:
2kλL/[y√(y² + L²/4)] in the direction away from the rod
Explanation:
For an element at x, the perpendicular component is dE_y = kλy dx/(x²+y²)^(3/2). Integrating from −L/2 to L/2 gives E_y = kλy[x/(y²√(x²+y²))] limits = 2kλL/[y√(y²+L²/4)].
7. A uniformly charged hemispherical shell of radius R occupies the region z ≥ 0 and has surface charge density σ. What is the electric field at the center due to the shell?
A.
πkσ in the positive z-direction
B.
2πkσ in the negative z-direction
C.
πkσ in the negative z-direction
D.
2πkσ in the positive z-direction
Correct Answer:
πkσ in the negative z-direction
Explanation:
For a surface element, dE = kσ dA/R². Its z-component at the center is dE_z = −kσ cosθ sinθ dθ dφ. Integrating θ from 0 to π/2 and φ from 0 to 2π gives E_z = −kσ(2π)(1/2) = −πkσ.
8. A uniformly charged quarter-circular arc of radius R lies in the first quadrant, extending from θ = 0 to θ = π/2, with linear charge density λ. What is the electric field at its center?
A.
(kλ/R)(î + ĵ)
B.
−(kλ/R)(î + ĵ)
C.
(2kλ/R)(î − ĵ)
D.
−(2kλ/R)î
Correct Answer:
−(kλ/R)(î + ĵ)
Explanation:
For dq = λR dθ, the field at the center is dE = (kλ/R)dθ directed opposite to the radial unit vector. Thus dE_x = −(kλ/R)cosθ dθ and dE_y = −(kλ/R)sinθ dθ. Both integrals from 0 to π/2 equal −kλ/R.
9. A circular disk of radius R has surface charge density σ(r) = σ₀r/R, where r is measured from its center. Find the electric field on its axis at a distance z > 0.
A.
σ₀/(2ε₀R) [√(z² + R²) − z]
B.
σ₀/(2ε₀R) [z − √(z² + R²)]
C.
σ₀R/(2ε₀) [1 − z/√(z² + R²)]
D.
σ₀/(2ε₀R) [z² + R² − z]
Correct Answer:
σ₀/(2ε₀R) [√(z² + R²) − z]
Explanation:
An annular element has dq = 2πrσ(r)dr = 2πσ₀r²dr/R. Its axial field is dE_z = k z dq/(z²+r²)^(3/2). Evaluating the resulting integral yields E_z = σ₀/(2ε₀R)[√(z²+R²)−z].
10. A uniformly charged circular arc of radius R subtends an angle 2α at its center and is symmetric about the positive x-axis. Its linear charge density is λ > 0. What is the electric field at the center?
A.
(2kλ/R)sinα in the positive x-direction
B.
(2kλ/R)cosα in the negative x-direction
C.
(2kλ/R)sinα in the negative x-direction
D.
(kλ/R)sin(2α) in the positive x-direction
Correct Answer:
(2kλ/R)sinα in the negative x-direction
Explanation:
Let the arc span θ = −α to α. Each element produces dE = kλ dθ/R directed inward toward the center. The transverse components cancel, while E_x = −(kλ/R)∫₋α^α cosθ dθ = −(2kλ/R)sinα.
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