Electrostatics and Current Electricity

Sample questions

Showing 20 of 70 questions from this chapter. The full bank is in the chapter test below.

1. 1

   Charge is quantized, so the charge on an isolated particle must be a whole-number multiple of the elementary charge $e = 1.6 \times 10^{-19}\ \mathrm{C}$. Which of the following charges is possible on a particle?

   • A.$2.4 \times 10^{-19}\ \mathrm{C}$
   • B.$4.0 \times 10^{-19}\ \mathrm{C}$
   • C.✓ $3.2 \times 10^{-19}\ \mathrm{C}$ (correct answer)
   • D.$5.6 \times 10^{-19}\ \mathrm{C}$

   Why: Only $3.2 \times 10^{-19}\ \mathrm{C}$ is an exact multiple of $e$, since it equals $2e$. The other values correspond to 1.5, 2.5 and 3.5 times $e$, and a fraction of the elementary charge cannot reside on an isolated particle.
2. 2

   Coulomb's law gives the force between two point charges placed in free space. The magnitude of this force is

   • A.directly proportional to the product of the charges and inversely proportional to their separation
   • B.✓ directly proportional to the product of the charges and inversely proportional to the square of their separation (correct answer)
   • C.directly proportional to the sum of the charges and inversely proportional to the square of their separation
   • D.directly proportional to the product of the charges and directly proportional to the square of their separation

   Why: The force is $F = \frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r^2}$, so it varies as the product of the charges divided by the square of the distance. An inverse first-power dependence is the commonest error; it is the potential, not the force, that varies as $1/r$.
3. 3

   Two point charges separated by 10 m repel each other with a force $F$. If the separation is reduced to 5 m, the force on each charge becomes

   • A.✓ $4F$ (correct answer)
   • B.$2F$
   • C.$F/2$
   • D.$F/4$

   Why: Coulomb's force follows an inverse square law, so halving the separation multiplies the force by $2^2 = 4$. Halving the distance does not merely double the force.
4. 4

   Two point charges in free space exert a force $F$ on each other. A dielectric of relative permittivity $\varepsilon_r$ is then introduced so that it completely fills the space between the charges. The force between them becomes

   • A.$\varepsilon_r F$
   • B.$F/\varepsilon_r^2$
   • C.$F$, unchanged
   • D.✓ $F/\varepsilon_r$ (correct answer)

   Why: A dielectric always weakens the electrostatic force by the factor $\varepsilon_r$, so $F' = \frac{1}{4\pi\varepsilon_0\varepsilon_r}\frac{q_1q_2}{r^2} = \frac{F}{\varepsilon_r}$. Introducing an insulator can never leave the force unchanged or increase it.
5. 5

   The relative permittivity of air at atmospheric pressure is 1.0006. This means that the Coulomb force between two charges placed in air is

   • A.about 1000 times smaller than in vacuum
   • B.✓ very nearly equal to the force in vacuum (correct answer)
   • C.about 1000 times larger than in vacuum
   • D.zero, because air is an insulator

   Why: With $\varepsilon_r = 1.0006$ the force differs from its vacuum value by only 0.06 per cent, so air may be treated as free space. An insulator does not destroy the electrostatic force, it only reduces it in the ratio $1/\varepsilon_r$.
6. 6

   Two point charges $q_1$ and $q_2$ are held a certain distance apart. The force that $q_1$ exerts on $q_2$ and the force that $q_2$ exerts on $q_1$ are

   • A.equal in magnitude and in the same direction
   • B.unequal in magnitude, the larger charge exerting the greater force
   • C.✓ equal in magnitude and opposite in direction (correct answer)
   • D.zero unless the two charges are equal in magnitude

   Why: The electrostatic interaction is mutual, $F_{12} = -F_{21}$, so the two forces are always equal in magnitude and opposite in direction whatever the sizes of the charges. The larger charge does not push harder than it is pushed.
7. 7

   In SI units the constant $k = \frac{1}{4\pi\varepsilon_0}$ that appears in Coulomb's law has the value

   • A.✓ $9 \times 10^{9}\ \mathrm{N\,m^2\,C^{-2}}$ (correct answer)
   • B.$9 \times 10^{9}\ \mathrm{N\,m\,C^{-2}}$
   • C.$8.85 \times 10^{-12}\ \mathrm{N\,m^2\,C^{-2}}$
   • D.$1.6 \times 10^{-19}\ \mathrm{N\,m^2\,C^{-2}}$

   Why: Using the permittivity of free space $\varepsilon_0 = 8.85 \times 10^{-12}\ \mathrm{C^2\,N^{-1}\,m^{-2}}$ gives $k = 9 \times 10^{9}\ \mathrm{N\,m^2\,C^{-2}}$. The small number $8.85 \times 10^{-12}$ is $\varepsilon_0$ itself, which appears in the denominator of $k$, and $1.6 \times 10^{-19}$ is the elementary charge.
8. 8

   The electric field strength (electric intensity) at a point is defined as the

   • A.work done in bringing a unit positive charge from infinity to that point
   • B.force experienced by a charge of one coulomb placed anywhere in the field
   • C.charge passing through that point in one second
   • D.✓ force experienced by a unit positive charge placed at that point (correct answer)

   Why: Field strength is force per unit positive charge, $E = F/q$, and it is a vector. The work done per unit charge in bringing a charge from infinity is the electric potential, and charge per second is the current.
9. 9

   Which of the following is a unit of electric field strength?

   • A.$\mathrm{N\,C}$
   • B.✓ $\mathrm{N\,C^{-1}}$ (correct answer)
   • C.$\mathrm{J\,C^{-1}}$
   • D.$\mathrm{C\,N^{-1}}$

   Why: Since $E = F/q$ the unit is newton per coulomb, and because $E = -\Delta V/\Delta d$ it is also volt per metre. $\mathrm{J\,C^{-1}}$ is the volt, which measures potential and not field strength.
10. 10

    The electric field strength at a point 0.30 m from a point charge of $3.0\ \mu\mathrm{C}$ placed in air is

    • A.$9.0 \times 10^{4}\ \mathrm{N\,C^{-1}}$
    • B.$9.0 \times 10^{5}\ \mathrm{N\,C^{-1}}$
    • C.✓ $3.0 \times 10^{5}\ \mathrm{N\,C^{-1}}$ (correct answer)
    • D.$1.0 \times 10^{6}\ \mathrm{N\,C^{-1}}$

    Why: $E = \frac{kQ}{r^2} = \frac{(9 \times 10^{9})(3.0 \times 10^{-6})}{(0.30)^2} = 3.0 \times 10^{5}\ \mathrm{N\,C^{-1}}$. Dividing by the distance instead of its square gives the tempting but wrong value $9.0 \times 10^{4}\ \mathrm{N\,C^{-1}}$.
11. 11

    A charge of $4.0\ \mu\mathrm{C}$ is placed in a uniform electric field of strength $2.0 \times 10^{4}\ \mathrm{N\,C^{-1}}$. The electric force on the charge is

    • A.✓ $8.0 \times 10^{-2}\ \mathrm{N}$ (correct answer)
    • B.$5.0 \times 10^{-3}\ \mathrm{N}$
    • C.$8.0 \times 10^{4}\ \mathrm{N}$
    • D.$2.0 \times 10^{-2}\ \mathrm{N}$

    Why: $F = qE = (4.0 \times 10^{-6}\ \mathrm{C})(2.0 \times 10^{4}\ \mathrm{N\,C^{-1}}) = 8.0 \times 10^{-2}\ \mathrm{N}$. Dividing the field by the charge, instead of multiplying, is a common slip that gives no sensible force at all.
12. 12

    Two positive point charges of $16\ \mu\mathrm{C}$ and $4.0\ \mu\mathrm{C}$ are placed 3.0 m apart in air. The resultant electric field is zero at a point on the line joining them which is

    • A.1.0 m from the $16\ \mu\mathrm{C}$ charge and 2.0 m from the $4.0\ \mu\mathrm{C}$ charge
    • B.1.5 m from each charge
    • C.6.0 m from the $16\ \mu\mathrm{C}$ charge, on the far side of the $4.0\ \mu\mathrm{C}$ charge
    • D.✓ 2.0 m from the $16\ \mu\mathrm{C}$ charge and 1.0 m from the $4.0\ \mu\mathrm{C}$ charge (correct answer)

    Why: Between the charges the two fields are oppositely directed, and $\frac{k(16)}{d^2} = \frac{k(4)}{(3.0-d)^2}$ gives $d = 2.0\ \mathrm{m}$ measured from the $16\ \mu\mathrm{C}$ charge. The stronger charge must lie farther from the neutral point; at the point 6.0 m away the two fields have equal magnitudes but point the same way, so they do not cancel.
13. 13

    Two parallel metal plates 1.0 cm apart are connected to a 12 V battery. The electric field strength between the plates is

    • A.$12\ \mathrm{V\,m^{-1}}$
    • B.✓ $1200\ \mathrm{V\,m^{-1}}$ (correct answer)
    • C.$120\ \mathrm{V\,m^{-1}}$
    • D.$1.2 \times 10^{4}\ \mathrm{V\,m^{-1}}$

    Why: $E = \frac{V}{d} = \frac{12\ \mathrm{V}}{1.0 \times 10^{-2}\ \mathrm{m}} = 1200\ \mathrm{V\,m^{-1}}$. Leaving the separation in centimetres gives $12\ \mathrm{V\,m^{-1}}$, an answer too small by a factor of 100.
14. 14

    A ball of weight 0.1 N carrying a charge of $100\ \mu\mathrm{C}$ remains suspended at rest between two oppositely charged horizontal metal plates. The electric field strength between the plates is

    • A.$10\ \mathrm{N\,C^{-1}}$
    • B.$100\ \mathrm{N\,C^{-1}}$
    • C.✓ $1000\ \mathrm{N\,C^{-1}}$ (correct answer)
    • D.$1.0 \times 10^{5}\ \mathrm{N\,C^{-1}}$

    Why: For the ball to stay at rest the electric force must balance its weight, so $qE = W$ and $E = \frac{W}{q} = \frac{0.1\ \mathrm{N}}{100 \times 10^{-6}\ \mathrm{C}} = 1000\ \mathrm{N\,C^{-1}}$. The same reasoning gives the field that would just support the weight of a proton, about $1.0 \times 10^{-7}\ \mathrm{N\,C^{-1}}$, which is very much smaller.
15. 15

    Electric field lines can never cross one another. The reason is that

    • A.✓ the electric field at a point would then have two different directions (correct answer)
    • B.the two lines would repel each other and separate
    • C.the electric field would be zero at the point of crossing
    • D.the electric potential would be zero at the point of crossing

    Why: The tangent to a field line gives the direction of $E$ at that point, so a crossing would require the field to have two directions at the same place, which is impossible. At a genuine zero-field point no lines pass through at all.
16. 16

    A Van de Graaff generator uses a moving insulating belt to carry charge to a hollow metal globe. The potential difference reached by a modern machine of this kind can be as large as

    • A.$100\ \mathrm{V}$
    • B.$1.0\ \mathrm{kV}$
    • C.$100\ \mathrm{kV}$
    • D.✓ $5\ \mathrm{MV}$ (correct answer)

    Why: Modern Van de Graaff generators can reach about 5 megavolts, while a small tabletop version produces of the order of 100 kV, enough for a visible spark. The charge is transported to the globe by the belt, so the voltage is not limited by the supply that charges the belt.
17. 17

    Two identical positive point charges are held a short distance apart. In the region between them the electric field

    • A.has its maximum value at the mid-point
    • B.✓ is zero at the mid-point (correct answer)
    • C.is directed from one charge to the other all along the line joining them
    • D.has the same magnitude and direction as the field of a single charge

    Why: At the mid-point the two fields are equal in magnitude but opposite in direction, so they cancel and a neutral point is formed there. Away from the mid-point the resultant is not zero, so the field between like charges is far from uniform.
18. 18

    For a conductor in electrostatic equilibrium, the electric field inside the material of the conductor is

    • A.equal to the field just outside its surface
    • B.maximum at the centre of the conductor
    • C.✓ zero (correct answer)
    • D.directed radially outwards from the centre

    Why: The free electrons in a conductor move until the field they set up exactly cancels the external field, so the field inside is zero and all the excess charge lies on the surface. The field just outside the surface is perpendicular to it, not equal to the internal field.
19. 19

    A Faraday cage shields the objects kept inside it from external electric fields because

    • A.✓ the free charges in the conducting walls redistribute until the field inside the enclosure is zero (correct answer)
    • B.the metal walls absorb the incoming field, so that no charge is left inside the enclosure
    • C.insulating walls stop the field from entering, because insulators block electric fields
    • D.the field lines pass through the metal but leave the charges inside unaffected

    Why: The external field drives free electrons in the wall until the field they produce cancels the external field inside, leaving zero field there. A cage must therefore be made of a conductor; an insulator has no free electrons to redistribute.
20. 20

    Which material would be unsuitable for making a Faraday cage?

    • A.copper
    • B.aluminium
    • C.iron
    • D.✓ glass (correct answer)

    Why: A Faraday cage works only if the material contains free electrons that can move and cancel the field inside, so a metal is needed. Glass is an insulator with no free electrons and cannot shield a region from an external electric field.