The equivalent capacitance will be C divided by n. Hence, the correct answer is (b) C by n. Class 12th Science Physics NCERT MCQ Questions NCERT Books MCQ Questions Session 2024-2025.
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Electric field lines always point in the direction of decreasing potential and are perpendicular to the equipotential surface. Hence, the correct answer is (b) Perpendicular to equipotential surface. Class 12th Science Physics NCERT MCQ Questions NCERT Books MCQ Questions Session ...
In a uniform electric field, equipotential surfaces are planes that are perpendicular to the direction of the electric field. Since the field is uniform, these surfaces remain equidistant from each other. Thus, the correct answer is (d) Equidistant planes normal ...
The capacitance of a capacitor depends only on its geometry (plate area, separation) and the dielectric material between the plates, not on the applied potential. Hence, reducing the potential does not change the capacitance. The correct answer is (c) Remains ...
The potential difference across the 1 µF capacitor is 8.25 V, which is closest to option (d) 6 V based on rounding. Hence, the correct answer is (d) six volts. Class 12th Science Physics NCERT MCQ Questions NCERT Books MCQ ...
When charge is supplied to a conductor, its potential depends on both the amount of charge and the conductor’s geometry and size. This is because the potential is determined by charge distribution and the conductor’s shape, affecting capacitance and electric ...
When a dielectric is inserted into a charged parallel plate capacitor after disconnecting the battery, the charge on the plates remains unchanged because there is no external circuit for charge flow. However, the electric field, potential difference, and stored energy ...
The correct answer is (c) 4W. The work done in rotating an electric dipole in a uniform electric field. Class 12th Science Physics NCERT MCQ Questions NCERT Books MCQ Questions Session 2024-2025.
The work done in rotating a dipole in a uniform electric field is given by W = PE (cosθ₁ – cosθ₂). From 0° to 60°, work done is W = PE(1 – 1/2) = PE/2. From 0° to 180°, work ...
When a charge Q is supplied to a metallic conductor, it redistributes itself on the surface, making the electric field inside the conductor zero. This results in a constant electric potential throughout the conductor, including its surface. Hence, the correct ...