NEET Physics Questions: Electrostatics

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A metallic solid sphere is placed in a uniform electric field. The lines of force follow the path(s) shown in figure as





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.
The protonic charge in 100 gm of water is________- c




Let q and 2q be the charges and r the distance between them Then $ 2 \times 10^{-3} = { 2 k q^2 \over r ^2 }$ and $ 5 \times 10^{-4} = { 2 kq^2 \over (r + 0.1 ) ^2 }$ $ so , {20 \over 5 } = { (r + 0.1) ^2 \over r^2 } $ $ \therefore 2 = { r +0.1 \over r^2 } \Rightarrow r = 0.1 m $ Now , substituting the value of r . So obtained 2
A copper sphere of mass 2 gm contains about $ 2 \times 10^{22} $ atoms. The charge on the nucleus of each atom is 29e. The fraction of electrons removed.




No explanation available.
The rate of alpha partical falls on neutral spheare is $10^{12} $ per second. The time in which sphere gets charged by 2 c is sec.




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 $2 ext{ C} / 3.2 imes 10^{-7} ext{ C/s} = 6.25 imes 10^6 ext{ s}$.
A charge Q is divided into two parts and then they are placed at a fixed distance. The force between the two charges is always maximum when the charges are .........




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 $Q/2$.
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Two small conducting sphere of equal radius have charges + 1 c and – 2 c respectively and placed at a distance d from each other experience force $F_1$. . If they are brought in contact and separated to the same distance, they experience force $F_2$. The ratio of $F_1 to F_2$ is ..........




No explanation available.
Two equal negative charges –q are fixed at points (o, a) and (o, –a). A positive charge Q is released from rest at the point (2a, o) on the X - axis. The charge Q will




No explanation available.
Four charges, each equal to –Q, are placed at the corners of a square and a charge +q is placed at its centre. If the system is in equilibrium, the value of q is




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 ) $.
Two identical charged spheres suspended from a common point by two massless strings of length l are initially a distance d (d << l ) apart because of their mutual repulsion. The charge begins to leak from both the spheres at a constant rate. As a result the spheres approach each other with a velocity Ο… . Then function of distance x between them becomes ...........




No explanation available.
Three identical spheres each having a charge q and radius R, are kept in such a way that each touches the other two spheares. The magnitude of the electric force on any sphere due to other two is ...........




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 $.
Two equal negative charges –q are fixed at points (o, a) and (o, –a) on the Y axis. A positive charge q is released from rest at the point x (x < < a) on the X-axis, then the frequency of motion




No explanation available.
A point charge q is situated at a distance r from one end of a thin conducting rod of length L having a charge Q (uniformly distributed along its length). The magnitude of electric force between the two, is ...............




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 $\frac{kqQ}{r(r+L)}$.
Two point charges of +16 c and –9 c are placed 8 cm apart in air distance of a point from –9 c charge at which the resultant electric field is zero.




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

Point charges 4 c and 2 c are placed at the vertices P and Q of a right angle triangle PQR respectively. Q is the right angle,$PR=2 \times10^{–2}m $ and $QR =10^{–2}m$ . The magnitude and direction of the resultant electric field at c is .........





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°

A small sphere whose mass is 0.1 gm carries a charge of $ 3 \times 10^{–10}C $ and is tieup to one end of a silk fibre 5 cm long. The other end of the fibre is attached to a large vertical conducting plate which has a surface charge of $ 25 \times 10^{–6}Cm^{–2}$ , on each side. When system is freely hanging the angle fibre makes with vertical is





No explanation available.