Q: The acceleration of an electron due to the mutual attraction between the electron and a proton when they are 1.6 Å apart is (m_e \approx 9 \times 10^{-31} kg, e = 1.6 \times 10^{-19} C, take \frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 N·m²·C⁻²)

10^{22} m/s². F = \frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{(1.6 \times 10^{-10})^2} = 9 \times 10^{-9} N a = \frac{F}{m_e} = \frac{9 \times 10^{-9}}{9 \times 10^{-31}} = 10^{22} m/s²

Q: Two pith balls carrying equal charges are suspended from a common point by strings of equal length, the equilibrium separation between them is r. Now, the strings are rigidly clamped at half the height. The equilibrium separation between the balls now becomes

\frac{r}{\sqrt[3]{2}}. For small-angle equilibrium, \tan\theta \approx \frac{r/2}{y} \propto \frac{Kq^2/r^2}{mg}, giving r^3 \propto y (string length). Halving the height (y → y/2) gives r'^3 = \frac{r^3}{2}, so r' = \frac{r}{2^{1/3}}

Question 1

The acceleration of an electron due to the mutual attraction between the electron and a proton when they are 1.6 Å apart is (m_e \approx 9 	imes 10^{-31} kg, e = 1.6 	imes 10^{-19} C, take \frac{1}{4\pi\varepsilon_0} = 9 	imes 10^9 N·m²·C⁻²)

Accepted Answer

10^{22} m/s². F = \frac{9 	imes 10^9 	imes (1.6 	imes 10^{-19})^2}{(1.6 	imes 10^{-10})^2} = 9 	imes 10^{-9} N

a = \frac{F}{m_e} = \frac{9 	imes 10^{-9}}{9 	imes 10^{-31}} = 10^{22} m/s²

Question 2

Two point charges A and B, having charges +Q and −Q respectively, are placed at certain distance apart and force acting between them is F. If 25% charge of A is transferred to B, then force between the charges becomes

Accepted Answer

\frac{9}{16}F. After transfer: Q_A' = \frac{3Q}{4}, Q_B' = -\frac{Q}{4}... magnitudes \frac{3Q}{4} and \frac{3Q}{4}.

F' = \frac{K \cdot \frac{3Q}{4} \cdot \frac{3Q}{4}}{r^2} = \frac{9}{16}\cdot\frac{KQ^2}{r^2} = \frac{9}{16}F

Question 3

Suppose the charge of a proton and an electron differ slightly. One of them is −e and the other is (e + Δe). If the net of electrostatic force and gravitational force between two hydrogen atoms placed at a distance d (much greater than atomic size) apart is zero, then Δe is of the order [mass of hydrogen m_h = 1.67 	imes 10^{-27} kg]

Accepted Answer

10^{-37} C. Setting electrostatic repulsion equal to gravitational attraction:

\frac{K(\Delta e)^2}{d^2} = \frac{Gm_h^2}{d^2}

(\Delta e)^2 = \frac{Gm_h^2}{K} = \frac{6.67 	imes 10^{-11} 	imes (1.67 	imes 10^{-27})^2}{9 	imes 10^9}

\Delta e \approx 1.44 	imes 10^{-37} C

Question 4

Two identical charged spheres suspended from a common point by two massless strings of lengths l, are initially at a distance d (d ≪ l) apart because of their mutual repulsion. The charges begin to leak from both the spheres at a constant rate. As a result, the spheres approach each other with a velocity v. Then, v varies as a function of the distance x between the spheres, as

Accepted Answer

v \propto x^{-1/2}. For small angles, equilibrium gives \frac{Kq^2}{x^2} = \frac{mgx}{2l}, so q^2 \propto x^3, i.e. q \propto x^{3/2}.

Since charge leaks at constant rate, \frac{dq}{dt} \propto x^{1/2}\cdot\frac{dx}{dt} is constant, so v = \frac{dx}{dt} \propto x^{-1/2}

Question 5

Two pith balls carrying equal charges are suspended from a common point by strings of equal length, the equilibrium separation between them is r. Now, the strings are rigidly clamped at half the height. The equilibrium separation between the balls now becomes

Accepted Answer

\frac{r}{\sqrt[3]{2}}. For small-angle equilibrium, 	an	heta \approx \frac{r/2}{y} \propto \frac{Kq^2/r^2}{mg}, giving r^3 \propto y (string length).

Halving the height (y → y/2) gives r'^3 = \frac{r^3}{2}, so r' = \frac{r}{2^{1/3}}

Question 6

Two positive ions, each carrying a charge q, are separated by a distance d. If F is the force of repulsion between the ions, the number of electrons missing from each ion will be (e being the charge on an electron)

Accepted Answer

\sqrt{\frac{4\pi\varepsilon_0 Fd^2}{e^2}}. With q = ne: F = \frac{1}{4\pi\varepsilon_0}\frac{n^2e^2}{d^2}

n = \sqrt{\frac{4\pi\varepsilon_0 Fd^2}{e^2}}

Question 7

An electron is moving round the nucleus of a hydrogen atom in a circular orbit of radius r. The Coulomb force \vec{F} between the two is (where k = \frac{1}{4\pi\varepsilon_0})

Accepted Answer

-k\frac{e^2}{r^3}\vec{r}. The force is attractive (directed towards the nucleus, i.e. along -\hat{r}):

\vec{F} = -k\frac{e^2}{r^2}\hat{r} = -k\frac{e^2}{r^3}\vec{r} (since \hat{r} = \frac{\vec{r}}{r})

Question 8

When air is replaced by a dielectric medium of constant K, the maximum force of attraction between two charges, separated by a distance

Accepted Answer

decreases K times. In a medium the force becomes F' = \frac{F}{K} — it decreases K times.

Question 9

A charge q is placed at the centre of the line joining two exactly equal positive charges Q. The system of three charges will be in equilibrium, if q is equal to

Accepted Answer

-\frac{Q}{4}. For each Q to be in equilibrium, the force from the other Q must balance the force from q. With charges Q at distance 2x and q at the centre (distance x):

\frac{Qq}{x^2} + \frac{Q^2}{(2x)^2} = 0 \Rightarrow q = -\frac{Q}{4}

Question 10

Point charges +4q, −q and +4q are kept on the x-axis at points x = 0, x = a and x = 2a, respectively. Then,

Accepted Answer

all the charges are in unstable equilibrium. The net force on each charge works out to zero, so all are in equilibrium. But displacing any charge produces a restoring-opposing force (e.g. displacing −q leads to a net attraction that increases the displacement), so the equilibrium is unstable for all.

Electric Charges and Coulomb's Law — NEET Physics PYQ MCQs with Solutions

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