NEET Practice Questions (MCQs) with Answers & Solutions

To calculate the time it takes for the sphere to get charged by 2 Coulombs with a rate of $10^{12}$ alpha particles per second, we need to know the charge of an alpha particle, which is $2 imes 1.6 imes 10^{-19}$ C. The total charge per second is $10^{12} imes 2 imes 1.6 imes 10^{-19} = 3.2 imes 10^{-7}$ C/s. Therefore, the time required to accumulate 2 C is $2 ext{ C} / 3.2 imes 10^{-7} ext{ C/s} = 6.25 imes 10^6 ext{ s}$.

The force between two charges is given by Coulomb's law: $F = \frac{k \cdot q_1 \cdot q_2}{r^2}$. To maximize the force, the product $q_1 \cdot q_2$ should be maximized. If the total charge Q is divided into two parts such that $q_1 = q$ and $q_2 = Q - q$, then $q_1 \cdot q_2 = q(Q - q)$. This product is maximized when $q = Q/2$. Therefore, the charges should be $Q/2$ and $Q/2$.

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