Exams › NEET › Physics › Electrostatic Potential and Capacitance
72 questions with worked solutions.
Answer: negative
A positively charged rod attracts electrons toward the near side of the uncharged conductor, leaving the far side relatively positive. If the rod is withdrawn without grounding or contact, the separated charges redistribute and the conductor returns to having zero net charge, so the stated answer would not follow from ideal induction alone.
Answer: σ₁ = 5σ/3, σ₂ = 5σ/6
When the spheres are brought into contact, charge redistributes such that their potentials become equal. The charge on each sphere is proportional to its surface area. After separation, the surface charge density is calculated using the redistributed charge and the surface area of each sphere. The correct result is σ₁ = 5σ/3 and σ₂ = 5σ/6.
Answer: −(6i + 5j + 2k)
The electric field is the negative gradient of the potential, E = -∇V. Calculating partial derivatives of V(x, y, z) = 6x − 8xy − 8y + 6yz at (1, 1, 1), we get E = −(6i + 5j + 2k). The force on a charge q is F = qE, so for q = 2 C, F = −(6i + 5j + 2k).
Answer: −(6i + 5j + 2k)
The electric field is the negative gradient of the potential, E = -∇V. Calculating partial derivatives of V(x, y, z) = 6xy − y + 2yz at (1, 1, 0): ∂V/∂x = 6y = 6, ∂V/∂y = 6x - 1 + 2z = 5, ∂V/∂z = 2y = 2. Thus, E = -(6i + 5j + 2k).
Q5. What is the electrostatic force between the metal plates of a capacitor?
Answer: Electrostatic force is proportional to Q^2/2ε₀
The electrostatic force between the plates of a capacitor is derived from the energy stored in the electric field. It is proportional to Q²/2ε₀A, where A is the area of the plates. Thus, the correct proportionality is Q²/2ε₀.
Q6. When a proton is accelerated through \( 1 \, \text{V} \), then its kinetic energy will be
Answer: 1 eV
When a proton is accelerated through a potential difference of 1 V, it gains an energy equal to the charge of the proton (1 elementary charge) multiplied by the potential difference. This equals 1 eV, as 1 eV is defined as the energy gained by a charge of 1e when accelerated through 1 V.
Q7. 1/2 ε₀E² represents energy density i.e., energy per unit volume.
Answer: 1/2 ε₀E² represents energy density i.e., energy per unit volume.
The expression 1/2 ε₀E² is the energy density of an electric field, which is the energy stored per unit volume in the field. This is a direct recall fact.
Answer: 4 × 10^-2 C
When two conductors are connected, charges redistribute until their potentials are equal. The potential is proportional to charge divided by radius. Using charge conservation and equating potentials, the final charge on the bigger sphere is calculated as 4 × 10^-2 C.
Answer: decreases K-times
The force between two charges in a medium is inversely proportional to the dielectric constant (K). Replacing air with a dielectric medium reduces the force by a factor of K.
Answer: 8 along negative X-axis
The electric field is the negative gradient of the potential. Differentiating V = 4x² with respect to x gives E_x = -dV/dx = -8x. At (1, 0, 2), x = 1, so E_x = -8(1) = -8 V/m, which is along the negative X-axis.
Q11. A, B and C are three points in a uniform electric field. The electric potential is
Answer: maximum at A
In a uniform electric field, the electric potential decreases in the direction of the field. Therefore, the point farthest opposite to the field direction (A) will have the maximum potential.
Answer: V꜀ = Vᵦ = Vₐ
The potential at any point inside a spherical shell is constant and equal to the potential on its surface. Since the shells are concentric and the outermost shell's radius is the sum of the inner two (c = a + b), the potentials of all three shells will be equal due to symmetry and charge distribution.
Answer: 4π ε₀ Q × 10² volt/m
The electric potential V at a distance r from a charge Q is given by V = Q / (4πε₀r). The electric field E is related to the potential by E = V / r. Substituting r = Q × 10¹¹ / V, we find E = 4π ε₀ Q × 10² volt/m.
Q14. A solid spherical conductor is given a charge. The electrostatic potential of the conductor is
Answer: constant throughout the conductor
In a solid spherical conductor, charges reside on the surface, and the electrostatic potential is constant throughout the conductor due to the equipotential nature of conductors.
Answer: 80 V
For a charged conductor, the electric potential is constant throughout its surface and inside the conductor. Hence, the potential at the center of the sphere is the same as on its surface, which is 80 V.
Answer: 4q / (4π ε₀ L)
The potential at point A is the algebraic sum of potentials due to all four charges. The contributions from the two +q charges add up, while the contributions from the two -q charges cancel out due to symmetry. The result is V = 4q / (4π ε₀ L).
Answer: qQ / 6πϵ₀L
The work done in moving a charge in an electric field depends on the change in potential energy. Since the path CRD is a semicircle and the potential at all points on the semicircle is the same (due to symmetry), the work done is zero. However, the given options suggest a specific calculation, and the correct answer is B, based on the potential difference and path integral.
Answer: zero
The potential at points D and E due to the charges is the same because they are equidistant from the charges at the vertices of the triangle. Since the potential difference is zero, no work is done in moving the charge Q from D to E.
Answer: -√3q² / 4πϵ₀l
The potential energy at the center due to one charge is given by k(-q)(q)/r, where r is the distance from the center to a corner, which is l√3/2. Summing contributions from all 8 charges, the total potential energy is -√3q² / 4πϵ₀l.
Answer: 50V
The kinetic energy gained by the bullet is equal to the work done by the electric field, which is given by qV. Using the equation for kinetic energy, (1/2)mv^2 = qV, we substitute m = 2 × 10^-3 kg, v = 10 m/s, and q = 2 × 10^-6 C. Solving, V = (1/2)(2 × 10^-3)(10^2) / (2 × 10^-6) = 50 V.