Practice free Electrostatics (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
A parallel plate capacitor has capacitance C. If it is equally filled with parallel layers of materials of dielectric constants K1 and K2 its capacity becomes C1. The ratio of C1 to C is
Two identical capacitors, have the same capacitance C. One of them is charged to potential V1 and the other to V2. The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is
Initial energy of the system
When the capacitors are joined, common potential
Final energy of the system
Decrease in energy =
Three capacitors of capacitance 3 μF are connected in a circuit. Then their maximum and minimum capacitances will be
,
A capacitor of capacity C1 is charged upto V volt and then connected to an uncharged capacitor of capacity C2. Then final potential difference across each will be
Common potential
Two identical thin rings each of radius R meters are coaxially placed at a distance R meters apart. If Q1 coulomb and Q2 coulomb are respectively the charges uniformly spread on the two rings, the work done in moving a charge q from the centre of one ring to that of other is
The work done in moving a charge q from the center of one ring to the other is given by the expression: W = (q(Q2 - Q1)(sqrt(2) - 1))/(sqrt(2)*4πε0R). This takes into account the attractive force between q and Q2 and the repulsive force between q and Q1.
A negatively charged plate has charge density of 2 × 10–6 C/m2. The initial distance of an electron which is moving toward plate but cannot strike the plate, if it is having energy of 200 eV
Consider two points 1 and 2 in a region outside a charged sphere. Two points are not very far away from the sphere. If E and V represent the electric field vector and the electric potential, which of the following is not possible
Outside the charged sphere, (for equal distances from centre) if electric fields at two points are same then both points must be equipotential points.
A uniform electric field pointing in positive x-direction exists in a region. Let A be the origin, B be the point on the x-axis at x = +1 cm and C be the point on the y-axis at y = +1 cm. Then the potentials at the points A, B and C satisfy
In a uniform electric field pointing in the positive x-direction, the electric potential increases as we move in the positive x-direction. Since point B (x = +1 cm) is farther along the positive x-axis than the origin A (x = 0), the potential at B is higher than the potential at A. Therefore, V_A < V_B.
The electric potential at a point (x, y) in the x – y plane is given by V = –kxy. The field intensity at a distance r from the origin varies as
Two equal point charges are fixed at x = –a and x = +a on the x-axis. Another point charge Q is placed at the origin. The change in the electrical potential energy of Q, when it is displaced by a small distance x along the x-axis, is approximately proportional to
When a point charge Q is displaced by a small distance x along the x-axis, the change in its potential energy due to the two fixed charges (-q at -a and +q at +a) is approximately proportional to x^2. This is because the potential energy of a point charge in an electric field varies inversely with the distance from the source charges.
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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.