Physics important questions

(a) Electric field intensity $\vec E$ at a point is the force per unit positive test charge placed there, $\vec E = \vec F/q$ (units $\text{N C}^{-1}$ or $\text{V m}^{-1}$). Potential gradient is the rate of change of potential with dist...

An erect, magnified image from a converging lens must be virtual, and that fixes the sign of the magnification, which we then feed into the lens formula. The lens is converging, so $f = +20\ \text{cm}$. Because the image is erect and thr...

We put the given distances into the lens formula, taking care with the signs, and then use the magnification to describe the image. A concave (diverging) lens has $f = -30\ \text{cm}$, and the real object gives $u = -20\ \text{cm}$. Rear...

An astronomical (refracting) telescope has a large, long-focus objective lens and a short-focus eyepiece. Parallel rays from a distant object are brought to a real, inverted image at the focus of the objective. This image lies at (or jus...

(a) The common defects of vision are myopia (short sight), hypermetropia (long sight), presbyopia and astigmatism.

In myopia, distant objects focus in front of the retina, so the far point lies closer than infinity. It is corrected with a diverging (concave) lens whose focal length equals the far-point distance, giving a power

$$P = -\frac{1}{x}$$

where $x$ is the far point.

In hypermetropia, near objects would focus behind the retina, so a converging (convex) lens is used to form a virtual image at the near point $N$ of an object placed at 25 cm, giving

$$\frac{1}{f} = \frac{1}{0.25} - \frac{1}{N}$$

(b) The Lensmaker's formula comes from applying refraction at the lens's two spherical surfaces (radii $R_1$ and $R_2$, refractive index $n$) in turn. At the first surface, light travels from air into glass, giving

$$\frac{n}{v_1} - \frac{1}{u} = \frac{n-1}{R_1}$$

At the second surface, the first image acts as the object as light passes back into air, giving

$$\frac{1}{v} - \frac{n}{v_1} = \frac{1-n}{R_2}$$

Adding the two equations eliminates $v_1$,

$$\frac{1}{v} - \frac{1}{u} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$

Since $\dfrac{1}{v}-\dfrac{1}{u} = \dfrac{1}{f}$ for a thin lens, this becomes

$$\boxed{\frac{1}{f} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)}$$

(a) The electric flux through a surface measures how many field lines cross it, defined as

$$\Phi = \oint \vec E\cdot d\vec A$$

Gauss's law states that the total flux out of any closed surface equals the enclosed charge divided by the permittivity of free space:

$$\oint \vec E\cdot d\vec A = \frac{q_{enc}}{\varepsilon_0}$$

To find the field of an infinite plane sheet carrying surface charge density $\sigma$, we take a cylindrical pillbox (each flat face of area $A$) piercing the sheet. By symmetry the field comes straight out of both faces, so the flux is $2EA$ while the charge enclosed is $\sigma A$:

$$2EA = \frac{\sigma A}{\varepsilon_0}$$

Cancelling $A$ gives

$$\boxed{E = \frac{\sigma}{2\varepsilon_0}}$$

which is independent of distance from the sheet.

(b) In a series combination every capacitor carries the same charge $Q$ while the voltages add up:

$$ \begin{aligned} V &= V_1 + V_2 + \dots \ &= \frac{Q}{C_1}+\frac{Q}{C_2}+\dots \end{aligned} $$

Dividing through by $Q$ gives the equivalent series capacitance,

$$\boxed{\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \dots}$$

In a parallel combination every capacitor has the same voltage $V$ while the charges add:

$$Q = C_1V + C_2V + \dots$$

so dividing by $V$ gives

$$\boxed{C_p = C_1 + C_2 + \dots}$$

Two lenses act in series here: the objective forms a small real image of the sun, and the eyepiece then projects that image, enlarged, onto the screen. By working back from the screen we can find the size of the small image, and from its size the angle the sun subtends.

We are given the objective focal length $f_o = 1\ \text{m}$, the eyepiece focal length $f_e = 2\ \text{cm}$, and the final real image of the sun, $H = 10\ \text{cm}$ tall, formed at $v = 24\ \text{cm}$ from the eyepiece. The objective forms its image at its focal plane, where the image height is $h_i = f_o,\alpha$ and $\alpha$ is the angle the sun subtends at the objective. That image is the object for the eyepiece, so once we know $h_i$ we can find $\alpha$.

For the eyepiece we use $\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$ with $f_e = 2\ \text{cm}$ and $v = 24\ \text{cm}$:

$$ \begin{aligned} \frac{1}{u} &= \frac{1}{2} - \frac{1}{24} \ &= \frac{11}{24} \end{aligned} $$

so $u = 2.18\ \text{cm}$, and the eyepiece magnification is

$$ \begin{aligned} m &= \frac{v}{u} \ &= \frac{24}{2.18} \ &= 11 \end{aligned} $$

The $10\ \text{cm}$ screen image is $m$ times the eyepiece's object, so the intermediate image formed by the objective was

$$ \begin{aligned} h_i &= \frac{H}{m} \ &= \frac{10}{11} \ &= 0.909\ \text{cm} \ &= 9.09\times10^{-3}\ \text{m} \end{aligned} $$

Using $h_i = f_o,\alpha$, the angle the sun subtends at the objective is

$$ \begin{aligned} \alpha &= \frac{h_i}{f_o} \ &= \frac{9.09\times10^{-3}}{1} \ &= 9.09\times10^{-3}\ \text{rad} \ &= 0.52^\circ \end{aligned} $$

Therefore the sun subtends about $9.1\times10^{-3}$ rad (about $0.52^\circ$) at the objective.

(a) Long sightedness, or hypermetropia, is a defect in which a person sees distant objects clearly but near objects appear blurred. The near point (the closest distance the eye can focus on) has shifted farther away than the normal 25 cm.

This happens because the eyeball is too short, so a sharp image would form behind the retina, or because the eye lens is too weak to converge the rays enough. Either way, rays from a near object are not brought to a focus on the retina.

The defect is corrected with a convex (converging) lens of suitable power placed in front of the eye. It bends the incoming rays inward a little, so the eye can now focus a near object on the retina and the near point is brought back to 25 cm.

(b) To derive the Lens maker's formula we consider refraction at the lens's two surfaces in turn. Take a thin lens of refractive index $\mu$ with surface radii $R_1$ and $R_2$.

At the first surface light travels from air into the glass, forming an intermediate image at distance $v_1$:

$$\frac{\mu}{v_1} - \frac{1}{u} = \frac{\mu - 1}{R_1}$$

At the second surface this intermediate image acts as the object as the light passes back into air:

$$\frac{1}{v} - \frac{\mu}{v_1} = \frac{1 - \mu}{R_2}$$

Adding the two equations eliminates $v_1$:

$$\frac{1}{v} - \frac{1}{u} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$

Since $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$ for a lens, we obtain the Lens maker's formula:

$$\boxed{\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)}$$

(a) Gauss's theorem states that the total flux through a closed surface equals $\dfrac{q{enc}}{\varepsilon0}$. For a plane charged conductor with surface charge density $\sigma$, the field just outside is normal to the surface. Take a sm...

The lens has a single focal length, so the two arrangements (before and after the lens is moved) must give the same $f$. Equating them lets us find the object distance and hence $f$. At first the screen (image) is $v1 = 30\ \text{cm}$ to...

Given: plate separation $d = 0.20\ \text{m}$, potential difference $V = 500\ \text{V}$; field uniform.

For a uniform field the potential gradient equals the field magnitude: $$ \begin{aligned} E &= \frac{V}{d} \ &= \frac{500}{0.20} \ &= 2500\ \text{V m}^{-1}. \end{aligned} $$ The electric field between the plates is $2.5\times10^{3}\ \text{V m}^{-1}$ (N C$^{-1}$), directed from the positive to the negative plate.

An electron of mass $m = 9.1\times10^{-31}\ \text{kg}$ and charge $q = 1.6\times10^{-19}\ \text{C}$ moves from rest through a uniform field $E = 1.2\times10^{4}\ \text{V/m}$ over a distance $s = 1\ \text{cm} = 0.01\ \text{m}$. The force on the electron is

$$ \begin{aligned} F &= qE \ &= (1.6\times10^{-19})(1.2\times10^{4}) \ &= 1.92\times10^{-15}\ \text{N} \end{aligned} $$

so its acceleration is

$$ \begin{aligned} a &= \frac{F}{m} \ &= \frac{1.92\times10^{-15}}{9.1\times10^{-31}} \ &= 2.11\times10^{15}\ \text{m/s}^2 \end{aligned} $$

Starting from rest, $s = \tfrac12 a t^2$, so the time taken is

$$ \begin{aligned} t &= \sqrt{\frac{2s}{a}} \ &= \sqrt{\frac{2(0.01)}{2.11\times10^{15}}} \ t &= \sqrt{9.48\times10^{-18}} \ &= 3.08\times10^{-9}\ \text{s} \end{aligned} $$

The electron takes about $3.08\times10^{-9}$ s (roughly 3.1 ns).

The charges are $+1\times10^{-6}\ \text{C}$ and $-4\times10^{-6}\ \text{C}$, separated by $2\ \text{m}$. For two unlike charges the null point (where $\vec E = 0$) lies outside the pair, on the side of the smaller charge. Let it be a dis...

(a) Chromatic aberration is the failure of a lens to bring different colours to a common focus. Because the refractive index of the glass varies with wavelength, violet light bends more than red and focuses nearer the lens, so coloured fringes appear around the image. It is a defect of lenses and is corrected by combining a convex and a concave lens of different glasses into an achromatic doublet.

Spherical aberration, on the other hand, is the failure of a spherical lens or mirror to bring edge (marginal) rays and central (paraxial) rays to the same focus. Rays through the edge focus closer than those near the axis. This occurs even for a single colour, and it is reduced by using a parabolic surface or by using a stop to block the marginal rays.

(b) The reason lies in the path the sunlight takes through the atmosphere, not the distance to the Sun. At noon the Sun is almost overhead, so its light travels down through a short, nearly vertical thickness of air. In the morning and evening the Sun is near the horizon, so its rays enter at a slant and pass through a far greater thickness of atmosphere. Over this longer path much more light is scattered and absorbed by air molecules and dust (Rayleigh scattering strongly removing the shorter wavelengths), so less intensity reaches the observer and the Sun looks dimmer and redder, even though its distance is essentially unchanged.

(a) Electrostatic shielding is the protection of a region from external electric fields by enclosing it in a conductor (a hollow conductor or cage). Since the field inside a conductor is zero, charges/fields outside cannot penetrate the cavity; this is why sensitive equipment is placed in a metal (Faraday) cage.

(b) On contact the charge redistributes until both reach the same potential. For two identical spheres the total charge shares equally, each getting $\dfrac{Q_1 + Q_2}{2}$. (For unequal conductors it distributes according to their capacitances to keep potentials equal.) If the charges were equal and opposite, they neutralise.

(a) A tree is tall and its moist, sappy interior conducts better than air, so it offers a preferred (low-resistance) path to ground: lightning tends to strike the tallest conductor. A person sheltering under it may be struck directly or ...

(a) For the parallel combination the total capacitance is $C_p = C_1 + C_2 + C_3$, whereas for the series combination

$$\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$$

which is always smaller than the smallest capacitor. So the parallel combination has the larger capacitance. For the series expression, the same charge $Q$ sits on each capacitor and the voltages add,

$$ \begin{aligned} V &= V_1 + V_2 + V_3 \ &= \frac{Q}{C_1} + \frac{Q}{C_2} + \frac{Q}{C_3} \end{aligned} $$

Since $V = Q/C_s$, dividing through by $Q$ gives

$$\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$$

(b) Gauss's law: the flux through a closed surface is

$$\oint\vec E\cdot d\vec A = \frac{q_{enc}}{\varepsilon_0}$$

For a plane charged conductor with surface charge density $\sigma$, take a cylindrical Gaussian pillbox with one flat face (area $A$) just outside the conductor and the other inside, where $E = 0$. Flux passes only through the outer face, so

$$E,A = \frac{\sigma A}{\varepsilon_0}$$

which gives

$$E = \frac{\sigma}{\varepsilon_0}$$

directed normally outward from the conductor's surface.

A capacitor $C1=6\ \mu\text{F}$ is charged to $V1=150\ \text{V}$, and its potential falls to a common value $V=90\ \text{V}$ when an initially uncharged capacitor $C2$ is connected in parallel. Charge is conserved when the two are joined...

In a series combination the two capacitors share the same charge, so we first find the equivalent capacitance, then the common charge, and finally the voltage across each.

With $C_1 = 4\ \mu\text{F}$ and $C_2 = 12\ \mu\text{F}$ in series, the equivalent capacitance comes from

$$ \begin{aligned} \frac{1}{C_s} &= \frac{1}{4} + \frac{1}{12} \ &= \frac{3+1}{12} \ &= \frac{4}{12} \end{aligned} $$

so $C_s = 3\ \mu\text{F}$. Across the $200\ \text{V}$ battery this stores a charge that is the same on each capacitor,

$$ \begin{aligned} Q &= C_s V \ &= 3\times10^{-6}\times200 \ &= 6\times10^{-4}\ \text{C} \ &= 600\ \mu\text{C} \end{aligned} $$

The potential difference across each is then this charge divided by its own capacitance. For the first,

$$ \begin{aligned} V_1 &= \frac{Q}{C_1} \ &= \frac{600\ \mu\text{C}}{4\ \mu\text{F}} \ &= 150\ \text{V} \end{aligned} $$

and for the second,

$$ \begin{aligned} V_2 &= \frac{Q}{C_2} \ &= \frac{600\ \mu\text{C}}{12\ \mu\text{F}} \ &= 50\ \text{V} \end{aligned} $$

As a check these add up to $150 + 50 = 200\ \text{V}$, the battery voltage. So each capacitor holds $600\ \mu\text{C}$, with $V_1 = 150\ \text{V}$ and $V_2 = 50\ \text{V}$.

The capacitor has $C = 245\times10^{-12}\ \text{F}$ with a charge $Q = 0.148\ \mu\text{C} = 0.148\times10^{-6}\ \text{C}$ on each plate, and the plate separation is $d = 5\ \text{mm} = 5\times10^{-3}\ \text{m}$. The potential difference ...

(a) As the vehicle moves, friction between the tyres, the body and the air generates static electric charge, which can accumulate to a high potential on the metal tank. A spark from this charge could ignite the inflammable vapour. The tr...

(a) The relative permittivity (dielectric constant) $\varepsilon_r$ tells us how many times the capacitance increases when the material fills the space between the plates compared with vacuum, so that $C = \varepsilon_r C_0$. Physically it measures how strongly the dielectric becomes polarised, which weakens the field between the plates and so lets the capacitor store more charge at the same voltage.

(b) To charge a body positively by induction, bring a negatively charged rod near (but not touching) the body, so the near face acquires induced positive charge and the far face negative. While the rod is held in place, earth the body momentarily by touching it, and electrons from the far (negative) face flow away to earth. Remove the earth connection first, then remove the rod. The body is left with a net positive charge, opposite in sign to the inducing rod.

(a) A rough surface has sharp points and edges where charge density and the surrounding field become very high (action of points), causing charge to leak away into the air by corona discharge. A highly polished surface has no such points...

The coefficient of linear expansion $\alpha$ is the fractional increase in length per unit rise in temperature,

$$\alpha = \frac{\Delta L}{L,\Delta\theta}$$

and the coefficient of cubical (volume) expansion $\gamma$ is the fractional increase in volume per unit rise in temperature,

$$\gamma = \frac{\Delta V}{V,\Delta\theta}$$

To relate them, take a cube of side $L$, so $V = L^3$. When heated, each side becomes $L(1+\alpha\Delta\theta)$, so the new volume is

$$V' = L^3(1+\alpha\Delta\theta)^3 \approx L^3(1 + 3\alpha\Delta\theta)$$

neglecting higher powers of the small quantity $\alpha\Delta\theta$. Comparing this with $V' = V(1+\gamma\Delta\theta)$ gives

$$\gamma = 3\alpha$$

For the numerical part, 10 litres of water has mass $m = 10\ \text{kg}$ (density $1000\ \text{kg m}^{-3}$), with $c = 4200\ \text{J kg}^{-1}\text{K}^{-1}$ and a temperature rise $\Delta\theta = 55 - 15 = 40\ \text{K}$. The heat required is

$$ \begin{aligned} Q &= mc,\Delta\theta \ &= 10 \times 4200 \times 40 \ &= 1.68\times10^{6}\ \text{J} \end{aligned} $$

So the heat required is $1.68\ \text{MJ}$.

(a) A conservative force does work that is independent of the path taken (zero over a closed loop), and this work can be stored as potential energy; gravity and the spring force are examples. A non-conservative force does path-dependent work and dissipates energy, friction being the usual example.

For the conservation of linear momentum, take two bodies that collide. By Newton's third law the mutual forces are equal and opposite and act for the same time $t$, so their impulses cancel when added, giving

$$m_1 v_1 + m_2 v_2 = m_1 u_1 + m_2 u_2$$

That is, the total momentum is unchanged when no external force acts.

(b) Hooke's law states that within the elastic limit stress is proportional to strain,

$$\text{stress} = Y\times\text{strain}$$

so the graph of load against extension is a straight line. When a wire is stretched the force grows from $0$ to $F$, so the work done is $W = \tfrac{1}{2}F e$. Dividing this by the volume $V = AL$ gives the energy density,

$$ \begin{aligned} u &= \frac{W}{AL} \ &= \frac{\tfrac{1}{2}Fe}{AL} \end{aligned} $$

which rearranges to

$$ \begin{aligned} u &= \frac{1}{2}\cdot\frac{F}{A}\cdot\frac{e}{L} \ &= \frac{1}{2},\text{stress}\times\text{strain} \end{aligned} $$

(c) The moment of inertia measures a body's resistance to a change in its rotational motion, $I = \sum m_i r_i^2$, and it depends on both the mass and how that mass is distributed about the axis. For a thin rod of mass $M$ and length $L$ about an axis through one end and perpendicular to its length, take the mass per unit length $\lambda = M/L$; an element $dx$ at distance $x$ contributes $dI = (\lambda,dx)x^2$. Integrating over the rod,

$$ \begin{aligned} I &= \int_0^L \lambda x^2,dx \ &= \lambda\frac{L^3}{3} \end{aligned} $$

and substituting $\lambda = M/L$,

$$ \begin{aligned} I &= \frac{M}{L}\cdot\frac{L^3}{3} \ &= \frac{ML^2}{3} \end{aligned} $$

(d) In a conical pendulum the bob moves in a horizontal circle, so the string sweeps out a cone at a fixed angle $\theta$, whereas a simple pendulum swings to and fro in a vertical plane. For a conical pendulum of string length $L$ and semi-vertical angle $\theta$, the vertical component of the tension balances the weight,

$$T\cos\theta = mg$$

while the horizontal component provides the centripetal force (with $r = L\sin\theta$),

$$ \begin{aligned} T\sin\theta &= \frac{mv^2}{r} \ &= m\omega^2 r \end{aligned} $$

Dividing the second relation by the first gives $\tan\theta = \dfrac{\omega^2 L\sin\theta}{g}$, so

$$\omega = \sqrt{\frac{g}{L\cos\theta}}$$

The time period and frequency then follow as

$$ \begin{aligned} \mathcal{T} &= 2\pi\sqrt{\frac{L\cos\theta}{g}} \ f &= \frac{1}{2\pi}\sqrt{\frac{g}{L\cos\theta}} \end{aligned} $$