I85PREP
⌘ K
Canary PAU Physics study materials
Canary PAU Physics study resources

Canary PAU Physics practice questions

One question at a time. Build your Canary PAU Physics understanding.

First 50 of 200 questions. Each one lists the choices, the correct answer, and an explanation. The checker at the top of this page saves progress in this browser when it is available.

Continue in the practice tool above to load more questions, 50 at a time.

Question 1Gravitational field and orbital mechanics

In Newton's law of universal gravitation, the force between two point masses is proportional to the product of the masses and inversely proportional to which quantity?

  1. The cube of the distance between them
  2. The distance between them
  3. The square root of the distance between them
  4. The square of the distance between them Correct answer.

Explanation

Newton's law states F = G·m1·m2/r², so the force falls off as the inverse square of the separation. Doubling the distance reduces the force to one quarter, which is the defining feature of an inverse-square law.

Question 2Gravitational field and orbital mechanics

The gravitational field strength at a point in space is defined as which of the following?

  1. The gravitational force acting on a body of unit weight at that point
  2. The gravitational force per unit mass experienced by a small test mass placed at that point Correct answer.
  3. The work needed to move a unit mass once around a closed orbit through that point
  4. The total gravitational potential energy of a unit mass at that point

Explanation

Field strength g is a vector defined as F/m, the force per unit mass on a test mass. Its units are N/kg, which are equivalent to m/s², consistent with it being a local acceleration.

Question 3Gravitational field and orbital mechanics

Which statement correctly describes the gravitational potential at a point in the field of an isolated point mass?

  1. It is a scalar whose value is zero at every point on an equipotential surface
  2. It is a positive scalar that is largest at the surface of the mass
  3. It is a vector pointing toward the mass and increasing with distance
  4. It is a scalar that is negative and tends to zero at infinity Correct answer.

Explanation

Gravitational potential V = -GM/r is a scalar, taken as zero at infinity, so it is negative everywhere at finite distance. It becomes less negative as r increases, approaching zero from below.

Question 4Gravitational field and orbital mechanics

A satellite moves in a circular orbit of radius r around a planet of mass M. Which expression gives its orbital speed?

  1. √(2GM/r)
  2. 2πr·√(GM/r)
  3. GM/r²
  4. √(GM/r) Correct answer.

Explanation

Setting the gravitational force equal to the centripetal requirement gives GMm/r² = mv²/r, so v² = GM/r and v = √(GM/r). The speed decreases as the orbital radius increases.

Question 5Gravitational field and orbital mechanics

Kepler's second law states that the line joining a planet to the Sun sweeps out equal areas in equal times. This is a direct consequence of which conservation principle?

  1. Conservation of mass
  2. Conservation of linear momentum
  3. Conservation of mechanical energy
  4. Conservation of angular momentum Correct answer.

Explanation

The gravitational force is central, so it exerts no torque about the Sun and the planet's angular momentum L = mr²ω is constant. Equal areas in equal times is exactly the statement that r²dθ/dt is constant.

Question 6Gravitational field and orbital mechanics

For a satellite in a circular orbit of radius r around a body of mass M, which expression gives the square of the orbital period T?

  1. T² = 4π²r³/(GM) Correct answer.
  2. T² = 2πr³/(GM)
  3. T² = GM/(4π²r³)
  4. T² = 4π²r/(GM)

Explanation

Combining v = √(GM/r) with T = 2πr/v gives T² = 4π²r³/(GM), which is Kepler's third law for circular orbits. The constant 4π²/(GM) depends only on the central mass.

Question 7Gravitational field and orbital mechanics

A probe is at a distance r from the centre of a planet of mass M. Which expression gives the minimum speed needed for the probe to escape the planet's gravity, ignoring other bodies?

  1. 2GM/r
  2. √(2GM/r) Correct answer.
  3. √(GM/r²)
  4. √(GM/r)

Explanation

Escape requires the total energy to be at least zero: ½mv² - GMm/r = 0, giving v = √(2GM/r). This is √2 times the circular orbital speed at the same radius.

Question 8Gravitational field and orbital mechanics

Two satellites orbit the same planet on circular paths with radii r and 4r. What is the ratio of their orbital periods (smaller radius to larger radius)?

  1. 1:4
  2. 1:64
  3. 1:2
  4. 1:8 Correct answer.

Explanation

Kepler's third law gives T ∝ r^(3/2), so a radius ratio of 1:4 produces a period ratio of 1^(3/2):4^(3/2) = 1:8. The period grows faster than the radius.

Question 9Gravitational field and orbital mechanics

A satellite in a circular orbit at radius r is given a small tangential boost, raising its speed above the circular value but below escape speed. What is the new orbit?

  1. A parabola opening away from the planet
  2. An ellipse with perigee at the boost point and apogee farther out Correct answer.
  3. A circle at a larger radius centred on the planet
  4. An ellipse with apogee at the boost point and perigee closer in

Explanation

Increasing the speed at a fixed point makes that point the closest approach of an elliptical orbit, so it becomes the perigee. The satellite then climbs to an apogee determined by its new energy, which is still negative.

Question 10Gravitational field and orbital mechanics

A planet of mass M and radius R has no atmosphere. Which expression gives the gravitational field strength at a height h above its surface?

  1. GM/(R + h)² Correct answer.
  2. GM/R² + gh
  3. GM/(R² + h²)
  4. GM/(R + h)

Explanation

Outside a spherically symmetric mass the field behaves as if all the mass were at the centre, so the distance from the centre is R + h. The inverse-square law then gives g = GM/(R + h)².

Question 11Gravitational field and orbital mechanics

Inside a uniform solid sphere of density ρ, at a distance r from the centre, the gravitational field strength is proportional to which quantity?

  1. r Correct answer.
  2. r²
  3. 1/r
  4. 1/r²

Explanation

Only the mass within radius r contributes to the field at that point, and that mass is proportional to r³. Dividing by r² leaves a field proportional to r, so it rises linearly from zero at the centre.

Question 12Gravitational field and orbital mechanics

A body of mass m is moved from the surface of a planet of mass M and radius R to a point very far away. Which expression gives the energy required, taking the potential as zero at infinity?

  1. GMm/(2R)
  2. 2GMm/R
  3. GMm/R²
  4. GMm/R Correct answer.

Explanation

The potential at the surface is -GM/R and at infinity is zero, so the change in potential energy is 0 - (-GMm/R) = GMm/R. This equals the kinetic energy needed for escape, consistent with ½mv² with v² = 2GM/R.

Question 13Gravitational field and orbital mechanics

Which statement about the equipotential surfaces of an isolated point mass is correct?

  1. They are radial lines passing through the mass
  2. They are concentric spheres, and the field lines run tangentially along them
  3. They are concentric spheres, and the field lines cross them at right angles Correct answer.
  4. They are parallel planes whose spacing is uniform

Explanation

Potential depends only on distance from the mass, so surfaces of constant potential are spheres centred on it. The field is the negative gradient of potential, so field lines are radial and perpendicular to these spheres.

Question 14Gravitational field and orbital mechanics

The gravitational potential at the surface of a planet is -GM/R. What is the potential at a point at height R above the surface?

  1. -GM/(4R)
  2. -GM/(2R) Correct answer.
  3. -2GM/R
  4. -GM/R

Explanation

At height R the distance from the centre is 2R, so the potential is -GM/(2R). The magnitude halves compared with the surface value, and the sign stays negative.

Question 15Gravitational field and orbital mechanics

A geostationary satellite stays above the same point on the equator. Which condition must its orbit satisfy?

  1. It must be polar and have a period equal to the planet's rotation period
  2. It must be equatorial and elliptical, with the planet at the nearer focus
  3. It must be equatorial and circular, with any period shorter than the rotation period
  4. It must be equatorial, circular, and have a period equal to the planet's rotation period Correct answer.

Explanation

To hover over a fixed equatorial point the satellite must orbit in the equatorial plane, in the same sense as the rotation, on a circle whose period matches one sidereal day. The matching period fixes a unique orbital radius.

Question 16Gravitational field and orbital mechanics

Two satellites of different masses orbit the same planet on the same circular path. How do their speeds and periods compare?

  1. The speeds are equal but the heavier satellite has a longer period
  2. Both the speeds and the periods are equal Correct answer.
  3. The heavier satellite moves faster and has a shorter period
  4. The lighter satellite moves faster and has a shorter period

Explanation

In GMm/r² = mv²/r the satellite mass cancels, so the orbital speed and period depend only on the central mass and the radius. Orbital motion is therefore independent of the orbiting body's mass.

Question 17Gravitational field and orbital mechanics

A planet moves in an elliptical orbit around the Sun. At which point is its speed greatest?

  1. The speed is the same at every point of the ellipse
  2. At the semi-minor axis endpoints, midway along the orbit
  3. At aphelion, the point farthest from the Sun
  4. At perihelion, the point closest to the Sun Correct answer.

Explanation

Angular momentum mr²ω is constant, so the angular speed is largest where r is smallest, at perihelion. Energy conservation gives the same result: potential energy is lowest there, so kinetic energy and speed are highest.

Question 18Gravitational field and orbital mechanics

The gravitational field strength at the surface of a planet is g and its radius is R. A rocket rises to a height equal to R above the surface. What is the field strength there?

  1. g/9
  2. g/4 Correct answer.
  3. 2g
  4. g/2

Explanation

At the surface g = GM/R², and at height R the distance from the centre is 2R, giving GM/(2R)² = g/4. The field falls to one quarter when the distance from the centre doubles.

Question 19Gravitational field and orbital mechanics

A satellite of mass m orbits at radius r around a body of mass M. Which expression gives its total mechanical energy?

  1. -GMm/(2r) Correct answer.
  2. -GMm/r
  3. GMm/(2r)
  4. -2GMm/r

Explanation

For a circular orbit the kinetic energy is GMm/(2r) and the potential energy is -GMm/r, so the total is -GMm/(2r). The total energy is negative, as required for a bound orbit, and equals half the potential energy.

Question 20Gravitational field and orbital mechanics

A satellite in a low circular orbit experiences a small drag force from the residual atmosphere. What happens to its orbit over time?

  1. It keeps the same radius but slows down
  2. It spirals inward while its speed increases Correct answer.
  3. It spirals outward while its speed decreases
  4. It escapes because the drag adds kinetic energy

Explanation

Drag removes mechanical energy, making the total energy more negative and lowering the orbital radius. Because the circular speed √(GM/r) rises as r falls, the satellite actually speeds up even though drag removes energy.

Question 21Gravitational field and orbital mechanics

A Hohmann transfer moves a spacecraft from a circular orbit of radius r1 to a larger circular orbit of radius r2. Which statement about the transfer is correct?

  1. One radial burn at r1 is enough to reach the higher circular orbit
  2. Two burns are used, and the transfer ellipse has apogee r1 and perigee r2
  3. No burns are needed because the spacecraft naturally spirals outward
  4. Two tangential burns are used, and the transfer ellipse has perigee r1 and apogee r2 Correct answer.

Explanation

A Hohmann transfer uses a first tangential burn to enter an ellipse tangent to both circular orbits, then a second burn at the far point to circularise. The ellipse has perigee at the lower orbit and apogee at the higher one.

Question 22Gravitational field and orbital mechanics

Which quantity is the same for every satellite in circular orbit at the same radius around a given planet?

  1. The orbital period Correct answer.
  2. The angular momentum of the satellite
  3. The gravitational force on the satellite
  4. The kinetic energy of the satellite

Explanation

The period follows from T² = 4π²r³/(GM), which contains only the central mass and the radius. Force, kinetic energy and angular momentum all depend on the satellite's own mass and therefore differ between satellites.

Question 23Gravitational field and orbital mechanics

The gravitational field strength at the surface of a planet is g and its radius is R. Which expression gives the escape speed from the surface?

  1. √(2gR) Correct answer.
  2. √(2g/R)
  3. 2gR
  4. √(gR)

Explanation

Escape speed is √(2GM/R), and since g = GM/R² we can write GM = gR², giving √(2gR²/R) = √(2gR). This is a convenient form because g and R are directly measurable.

Question 24Gravitational field and orbital mechanics

A planet has the same mass as the Earth but twice the radius. How does the gravitational field strength at its surface compare with that at the Earth's surface?

  1. It is four times as large
  2. It is one half as large
  3. It is one quarter as large Correct answer.
  4. It is twice as large

Explanation

Surface field strength is GM/R², so at fixed mass doubling the radius divides the field by four. The planet is less dense and its surface gravity is weaker.

Question 25Gravitational field and orbital mechanics

A binary system consists of two stars of masses M and 3M separated by a distance d. Where is the centre of mass of the system?

  1. At a distance d/4 from the centre of the more massive star Correct answer.
  2. At a distance 3d/4 from the centre of the more massive star
  3. At the centre of the more massive star
  4. At a distance d/2 from the centre of the more massive star

Explanation

The centre of mass divides the separation in inverse proportion to the masses, so the 3M star is three times closer to it than the M star. The distance from the 3M star is therefore d·M/(M + 3M) = d/4.

Question 26Gravitational field and orbital mechanics

A spacecraft moves along a path where the gravitational potential changes by +2.0 × 10^7 J/kg between two points. The spacecraft has mass 800 kg. What is the change in its gravitational potential energy?

  1. +4.0 × 10^4 J
  2. -1.6 × 10^10 J
  3. +2.5 × 10^4 J
  4. +1.6 × 10^10 J Correct answer.

Explanation

Gravitational potential energy is U = mφ, so a change in potential Δφ gives ΔU = mΔφ. Multiplying 800 kg by +2.0 × 10^7 J/kg yields +1.6 × 10^10 J. The positive sign is consistent with moving to a region of higher (less negative) potential.

Question 27Gravitational field and orbital mechanics

A moon orbits a planet on a circular path of radius r with speed v. Which expression gives the mass of the planet?

  1. vr²/G
  2. Gr/v²
  3. v²r/G Correct answer.
  4. v²/(Gr)

Explanation

For a circular orbit the gravitational force supplies the centripetal force, so GMm/r² = mv²/r. Cancelling m and one factor of r gives GM = v²r, hence M = v²r/G. The moon's own mass cancels, so the result depends only on the orbit.

Question 28Gravitational field and orbital mechanics

Which statement about the gravitational field lines around an isolated uniform sphere is correct?

  1. Outside the sphere the lines are radial and their spacing decreases with distance.
  2. Outside the sphere the lines are radial and their spacing increases with distance. Correct answer.
  3. Outside the sphere the lines form closed loops returning to the sphere.
  4. Outside the sphere the lines are radial and equally spaced at all distances.

Explanation

Field lines point toward the centre of mass and are radial by symmetry. Their density represents field strength, which falls as 1/r², so the lines spread apart as distance increases. Closed loops are characteristic of magnetic fields, not gravitational ones.

Question 29Gravitational field and orbital mechanics

A satellite in a circular orbit around a planet has period T and orbital radius r. A second satellite orbits the same planet with period 8T. What is its orbital radius?

  1. 2r
  2. 4r Correct answer.
  3. 64r
  4. 8r

Explanation

Kepler's third law states that T² is proportional to r³ for orbits around the same central body. If the period increases by a factor of 8, then r³ increases by 8² = 64, so r increases by the cube root of 64, which is 4. Hence the new radius is 4r.

Question 30Gravitational field and orbital mechanics

A comet moves on an elliptical orbit around the Sun. At perihelion its distance is 0.5 AU and its speed is 40 km/s. At aphelion its distance is 4.0 AU. What is its speed at aphelion?

  1. 5 km/s Correct answer.
  2. 8 km/s
  3. 20 km/s
  4. 10 km/s

Explanation

Conservation of angular momentum for a central force gives r·v constant at perihelion and aphelion, where velocity is perpendicular to the radius. Thus 0.5 × 40 = 4.0 × v, giving v = 5 km/s. The comet moves slowest at its greatest distance from the Sun.

Question 31Gravitational field and orbital mechanics

Two point masses, 6.0 kg and 4.0 kg, are separated by 0.50 m. At what distance from the 6.0 kg mass is the net gravitational field zero?

  1. 0.30 m
  2. 0.20 m
  3. 0.27 m Correct answer.
  4. 0.23 m

Explanation

At the null point the field strengths from the two masses are equal in magnitude and opposite in direction: GM1/x² = GM2/(d−x)². Taking the square root gives (d−x)/x = √(M2/M1) = √(4/6) ≈ 0.816, so x ≈ 0.50/1.816 ≈ 0.27 m. The null point lies closer to the smaller mass.

Question 32Gravitational field and orbital mechanics

A planet has mass M and radius R. A projectile is launched vertically from the surface with speed equal to half the escape speed. Ignoring air resistance and rotation, what maximum height above the surface does it reach?

  1. 3R/4
  2. R
  3. R/3 Correct answer.
  4. R/2

Explanation

Energy conservation gives ½v² = GM(1/R − 1/r) with v = ½v_esc = ½√(2GM/R), so ½v² = GM/(4R). Then 1/R − 1/r = 1/(4R), giving 1/r = 3/(4R) and r = 4R/3. The height above the surface is r − R = R/3.

Question 33Gravitational field and orbital mechanics

A satellite of mass m is in a circular orbit of radius r around a planet of mass M. Which expression gives the magnitude of the additional energy needed to move it to a circular orbit of radius 2r?

  1. GMm/(4r) Correct answer.
  2. 3GMm/(4r)
  3. GMm/r
  4. GMm/(2r)

Explanation

The total energy of a circular orbit is E = −GMm/(2r). The change is ΔE = −GMm/(4r) − (−GMm/(2r)) = +GMm/(4r). The positive sign confirms that raising the orbit requires energy input.

Question 34Gravitational field and orbital mechanics

A probe is in a circular orbit of radius r around a planet. It fires its engine briefly in the direction of motion and enters an elliptical transfer orbit whose apocentre is at 3r. Which statement about the transfer orbit is correct?

  1. Its semimajor axis is 2r and its period is greater than the original circular period. Correct answer.
  2. Its semimajor axis is 3r and its period is greater than the original circular period.
  3. Its semimajor axis is 1.5r and its period is less than the original circular period.
  4. Its semimajor axis is 2r and its period is equal to the original circular period.

Explanation

The transfer ellipse has pericentre r and apocentre 3r, so its major axis is 4r and its semimajor axis is 2r. By Kepler's third law, T² ∝ a³, so a larger semimajor axis means a longer period. The period of the ellipse exceeds that of the original circle of radius r.

Question 35Gravitational field and orbital mechanics

A planet of mass M and radius R rotates with angular speed ω. Which condition must hold for a loose object at the equator to remain on the surface without being thrown off?

  1. ω²R² ≤ GM/R
  2. ω²R ≤ GM/R² Correct answer.
  3. ω²R ≥ GM/R²
  4. ωR ≤ GM/R²

Explanation

At the equator the required centripetal acceleration is ω²R, supplied by gravity minus the normal force. For the object to stay in contact, gravity must be at least as large as the required centripetal acceleration, so ω²R ≤ GM/R². Equality defines the maximum rotation rate before material is lost.

Question 36Gravitational field and orbital mechanics

A satellite is in a circular orbit at radius r. Which graph shape would correctly represent how the square of its period varies with r³ for orbits around a single central body?

  1. A curve whose slope decreases as r³ increases
  2. A straight line with a positive intercept on the T² axis
  3. A straight line through the origin with slope GM/(4π²)
  4. A straight line through the origin with slope 4π²/(GM) Correct answer.

Explanation

Kepler's third law in Newtonian form is T² = (4π²/GM)r³, so a plot of T² against r³ is linear. The line passes through the origin because r → 0 gives T → 0 in this idealised relation. The slope depends only on the central mass.

Question 37Gravitational field and orbital mechanics

A planet of mass M and radius R has a uniform density. Which expression gives the gravitational potential at a distance r = R/2 from its centre?

  1. 0
  2. −GM(3R² − r²)/(2R³) evaluated at r = R/2 Correct answer.
  3. −GM/R
  4. −GM/r evaluated at r = R/2

Explanation

Inside a uniform sphere the potential is φ = −GM(3R² − r²)/(2R³), which reduces to −GM/R at the surface. Substituting r = R/2 gives −GM(3R² − R²/4)/(2R³) = −11GM/(8R). Simply using −GM/r would incorrectly treat the sphere as a point mass.

Question 38Gravitational field and orbital mechanics

A spacecraft is in a circular orbit of radius r around a planet. It makes a small tangential burn that increases its speed by a factor of 1.05 while its position is unchanged. Which statement best describes the new orbit?

  1. The spacecraft escapes the planet because any increase above circular speed causes escape.
  2. The orbit becomes an ellipse with the burn point as pericentre and a semimajor axis larger than r. Correct answer.
  3. The orbit remains circular because the speed increase is small.
  4. The orbit becomes an ellipse with the burn point as apocentre and a semimajor axis smaller than r.

Explanation

At the burn point the speed exceeds the circular value but remains below escape speed, so the point becomes the pericentre of an ellipse. The new total energy is less negative than before, so the semimajor axis increases. A tangential boost always raises the opposite side of the orbit.

Question 39Gravitational field and orbital mechanics

A binary system consists of two stars of masses M and 2M separated by a distance d. They orbit their common centre of mass. Which expression gives the orbital period of the system?

  1. 2π√(d³/(3GM)) Correct answer.
  2. 2π√(d³/(GM))
  3. 2π√(d³/(2GM))
  4. 2π√(3d³/(GM))

Explanation

For a two-body system the relative motion obeys Kepler's third law with the total mass: T² = 4π²d³/(G(M1 + M2)). Here the total mass is 3M, so T = 2π√(d³/(3GM)). Using only one star's mass would give the wrong period.

Question 40Gravitational field and orbital mechanics

A satellite orbits a planet at radius r with speed v. Which statement correctly describes the relationship between the satellite's kinetic energy and its gravitational potential energy?

  1. The kinetic energy equals twice the magnitude of the potential energy.
  2. The kinetic energy equals the magnitude of the potential energy.
  3. The kinetic energy equals one quarter of the magnitude of the potential energy.
  4. The kinetic energy equals half the magnitude of the potential energy. Correct answer.

Explanation

For a circular orbit, mv²/r = GMm/r² gives mv² = GMm/r, so K = ½mv² = GMm/(2r). The potential energy is U = −GMm/r, whose magnitude is GMm/r. Therefore K = |U|/2, a result known as the virial relation for circular orbits.

Question 41Electric field and electrostatics

Two point charges are placed a distance r apart in vacuum. If the distance between them is doubled while the charges are unchanged, by what factor does the magnitude of the electrostatic force between them change?

  1. It becomes four times its original value
  2. It doubles
  3. It becomes one half of its original value
  4. It becomes one quarter of its original value Correct answer.

Explanation

Coulomb's law states that the force is inversely proportional to the square of the separation, F = kq1q2/r². Doubling r multiplies the force by 1/2² = 1/4. The charges are unchanged, so only the geometric factor matters.

Question 42Electric field and electrostatics

Which statement correctly describes the electric field vector at a point in space?

  1. It is the force per unit negative charge placed at that point
  2. It is the potential energy per unit charge at that point
  3. It is the force per unit positive charge placed at that point Correct answer.
  4. It is the total charge enclosed by a surface through that point

Explanation

The electric field is defined as E = F/q for a small positive test charge, so it points in the direction of the force on a positive charge. Potential energy per unit charge defines electric potential, not field. The enclosed charge relates to flux through a surface, not the field at a point.

Question 43Electric field and electrostatics

The electric field lines around an isolated positive point charge are best described as:

  1. Closed loops encircling the charge
  2. Radial lines pointing outward from the charge Correct answer.
  3. Parallel lines of constant spacing
  4. Radial lines pointing inward toward the charge

Explanation

Field lines indicate the direction of the force on a positive test charge, which is repelled away from a positive source. Hence the lines radiate outward. Inward radial lines would describe a negative charge, and closed loops are characteristic of magnetic fields.

Question 44Electric field and electrostatics

The electric potential at a distance r from a point charge Q is V = kQ/r. What is the potential at a point where the electric field due to the same charge has magnitude E?

  1. V = E/r
  2. V = Er Correct answer.
  3. V = E/r²
  4. V = E r²

Explanation

For a point charge E = kQ/r² and V = kQ/r, so dividing gives V/E = r, hence V = Er. This relation holds for the radial field of a point charge. It is a useful shortcut when only E and r are known.

Question 45Electric field and electrostatics

A charge of +2 μC is placed in a uniform electric field of 500 N/C directed to the right. What is the magnitude and direction of the electric force on the charge?

  1. 2.5×10⁻⁹ N to the right
  2. 1×10⁻³ N to the right Correct answer.
  3. 4×10⁻⁴ N to the left
  4. 1×10⁻³ N to the left

Explanation

The force magnitude is F = qE = (2×10⁻⁶ C)(500 N/C) = 1×10⁻³ N. Because the charge is positive, the force is parallel to the field, i.e. to the right. Sign of the charge determines the direction relative to E.

Question 46Electric field and electrostatics

Which of the following is a correct statement of the principle of conservation of electric charge?

  1. The total charge of an isolated system remains constant Correct answer.
  2. Charge is conserved only in vacuum, not in conductors
  3. The total charge always decreases over time in any system
  4. Charge can be created whenever a conductor is rubbed

Explanation

In an isolated system the net charge cannot change; charge may be transferred between bodies but not created or destroyed. Rubbing simply moves electrons from one body to another, leaving the total unchanged. Conservation applies in all media.

Question 47Electric field and electrostatics

Two identical small conducting spheres carry charges +6 μC and −2 μC. They are brought into contact and then separated. What is the charge on each sphere?

  1. +2 μC on each Correct answer.
  2. +6 μC and −2 μC, unchanged
  3. −2 μC on each
  4. +4 μC on each

Explanation

For identical conductors in contact, the total charge is shared equally. The total is (+6) + (−2) = +4 μC, so each sphere receives +2 μC. The charges are identical because the spheres are identical.

Question 48Electric field and electrostatics

The SI unit of electric field strength can be expressed as:

  1. C/N, equivalent to V·m
  2. N/C, equivalent to V/m Correct answer.
  3. N·m²/C²
  4. J/C, equivalent to V

Explanation

Field is force per unit charge, so N/C. Since potential difference is work per unit charge, V/m is dimensionally identical to N/C. The other options correspond to inverse field, potential, and the Coulomb constant.

Question 49Electric field and electrostatics

A point charge of 4×10⁻⁹ C is placed at the origin. What is the magnitude of the electric field at a point 0.2 m away, taking k = 9×10⁹ N·m²/C²?

  1. 1.8×10² N/C
  2. 3.6×10² N/C
  3. 9×10² N/C Correct answer.
  4. 4.5×10² N/C

Explanation

Using E = kQ/r² = (9×10⁹)(4×10⁻⁹)/(0.2)² = 36/0.04 = 900 N/C. The calculation uses the square of the distance in metres. The result is 9×10² N/C.

Question 50Electric field and electrostatics

Two charges, +q and −q, are separated by a distance d. At the midpoint of the line joining them, the electric field is:

  1. Non-zero and directed toward the positive charge
  2. Zero because the contributions cancel
  3. Zero because the potential also vanishes there
  4. Non-zero and directed toward the negative charge Correct answer.

Explanation

At the midpoint the field from the positive charge points away from it (toward the negative charge), and the field from the negative charge also points toward the negative charge. The two contributions add, giving a non-zero field toward the negative charge. The potential, by contrast, cancels to zero at the midpoint.