Electrostatics MCQs for NEET — Physics Questions with Answers

When a metallic solid sphere is placed in a uniform electric field, the lines of electric force are distorted due to the conducting nature of the sphere. The field inside the sphere is zero, and the field lines are perpendicular to the surface of the sphere. The correct representation of this phenomenon is shown in option 4, where the lines of force bend around the sphere and are perpendicular at the points where they touch the sphere.

In order for the system to be in equilibrium, the net force on the charge +q at the center must be zero. The four charges extendash Q at the corners of the square create electric fields that cancel each other out at the center. However, the charge +q at the center will experience a force due to each of the -Q charges. The condition for equilibrium dictates that the force due to the central charge must balance the combined forces of the corner charges. Calculating this, we find the correct value of q to be $ - { Q extbackslash over 4 } ( 1 + 2 extbackslash sqrt 2 ) $.

When three identical charged spheres each with charge q and radius R are kept such that they touch each other, the distance between the centers of any two spheres is 2R. The force between any two charges is given by Coulomb's law: $ F = extbackslash frac{1}{4 extbackslash pi extbackslash epsilon_0} extbackslash frac{q^2}{(2R)^2} $. Since there are two such forces acting at an angle of 120 degrees with respect to each other, the resultant force can be calculated using vector addition. The correct expression for this combined force is $ extbackslash frac{1}{4 extbackslash pi extbackslash epsilon_0} extbackslash frac{ extbackslash sqrt 3}{4} extbackslash left ( extbackslash frac{q}{R} extbackslash right ) ^2 $.

To determine the force between a point charge and a uniformly charged rod, we use the principle of superposition and Coulomb's law. Integrating the contributions of small charge elements along the length of the rod gives the expression for the force. The correct formula is $\frac{kqQ}{r(r+L)}$, where k is the Coulomb constant, q is the point charge, Q is the total charge on the rod, and r and L are the given distances. Hence, the correct option is $\frac{kqQ}{r(r+L)}$.

$ { k . q_1 \over (x + 0.08 ) ^2 } = { kq_2 \over x^2 } $ then find x

EP = [k(4 × 10–6)/(PR)2] = [(9 × 109 × 4 × 10–6)/(4 × 10–4)] = 9 ×107 N/C EQ = [k(2 × 10–6)/(QR)2] = [(9 × 109 × 2 × 10–6)/(10–4)] = 18 × 107 N/C in Δ PQR, cos θ = [(10–2)/(2 × 10–2)] = (1/2) i.e. θ = 60° E = √(EP2 + EQ2 + 2EPEQ cos 60) = √[(9 × 107)2 + (18 × 107)2 + {2 × 9 × 107 × 18 × 107 × (1/2)}] = √(81 × 1014 + 324 × 1014 + 162 × 1014) = √(547 × 1014) = 2.38 × 108 N/C tan α = [(EQ sin θ)/(EP + EQ cos θ)] = [(18 × 107 sin 60)/{(9 × 107 + 18 × 107 × cos 60)] = [{(18 × 107 × (√3/2)}/(9 × 107 + 9 × 107)] tan α = (√3/2) α = 40.89°