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Unit notes and practice questions

Grade 11 Physics unit 6: Electrostatics and Electric Circuit

Everything here is for unit 6 of the Grade 11 Physics textbook: what the chapter covers, the short notes, practice questions with their answers, and the lesson video where one exists.

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Practice questions on unit 6
217–286
The textbook pages for unit 6

Short notes for unit 6

The cards filed under this chapter on the Grade 11 Physics page. Open one at a time.

F=k∣q1q2∣r2F = k\frac{|q_1 q_2|}{r^2}
FF
magnitude of the electrostatic force between the two point chargesNN
kk
Coulomb constant, k = 1/(4 pi epsilon_0) = 9.0 x 10^9N m2 C−2N\,m^2\,C^{-2}
q1,q2q_1, q_2
the two point chargesCC
rr
distance between the two chargesmm

When you use it

Use this when you need the electrostatic force between two point charges separated by a distance r.

Watch out

Convert the distance to metres before squaring it. For example 1.0 cm = 0.01 m, so r^2 = 1.0 x 10^-4 m^2, not 1.0. Attraction or repulsion comes from the signs of the charges, not from the formula.

Drafted from Grade 11 Physics, pages 217-286, then checked twice before it went up

E=kQr2E = k\frac{Q}{r^2}
EE
electric field strength at distance r from the point chargeN C−1N\,C^{-1}
kk
Coulomb constantN m2 C−2N\,m^2\,C^{-2}
QQ
source point charge that produces the fieldCC
rr
distance from the source charge to the point of interestmm

When you use it

Use this when you need the electric field strength at a point a distance r from a single point charge.

Watch out

The field does not depend on a test charge, so do not multiply by q. For several charges, add the field vectors with their directions, not their magnitudes. Field points away from a positive source and towards a negative source.

Drafted from Grade 11 Physics, pages 217-286, then checked twice before it went up

R=VIR = \frac{V}{I}
RR
resistance of the conductor or resistorΩ\Omega
VV
potential difference across the conductorVV
II
current through the conductorAA

When you use it

Use this to find resistance, voltage or current in a metal conductor when its temperature stays constant.

Watch out

Ohm's law holds only at constant temperature. A filament lamp is non-ohmic: its resistance rises as it heats, so its V-I graph curves rather than being a straight line through the origin.

Drafted from Grade 11 Physics, pages 217-286, then checked twice before it went up

Practice questions on unit 6

10 multiple-choice questions written against this chapter, each with its answer and the reason. Try one before you open the answer.

Question 1

According to the quantization of electric charge, how many electrons are needed to make a total charge of -1.0 C?

  1. A.6.25 x 10^18 electrons
  2. B.1.60 x 10^19 electrons
  3. C.6.25 x 10^-19 electrons
  4. D.1.60 x 10^-18 electrons
Show the answer and why

Answer: A 6.25 x 10^18 electrons

Why

Charge is quantized as q = ne. Dividing the charge 1.0 C by the elementary charge 1.6 x 10-19 C gives 6.25 x 1018 electrons.

Question 2

Which of the following is a correct property of electric field lines?

  1. A.They cross each other at right angles.
  2. B.They begin on positive charges and end on negative charges.
  3. C.They are parallel to the surface of a charged conductor.
  4. D.Farther spaced lines indicate a stronger electric field.
Show the answer and why

Answer: B They begin on positive charges and end on negative charges.

Why

Electric field lines always begin on positive charges and radiate outward toward negative charges, where they terminate. They never cross each other and are always perpendicular to the surface of a charged body.

Question 3

Which of the following pairs of units are both valid units for electric field strength?

  1. A.N/C and V/m
  2. B.J/C and V/m
  3. C.N/m and V/C
  4. D.A/m and N/C
Show the answer and why

Answer: A N/C and V/m

Why

Electric field strength is defined as force per unit charge (N/C), which is also mathematically equivalent to potential difference per unit distance (V/m): 1 N/C = 1 V/m.

Question 4

A headlamp in a vehicle draws a current of 2.50 A when a potential difference of 12.0 V is applied across it. What is its resistance?

  1. A.0.21 ohms
  2. B.4.8 ohms
  3. C.14.5 ohms
  4. D.30.0 ohms
Show the answer and why

Answer: B 4.8 ohms

Why

Using Ohm's law, R = V / I. Substituting the values gives R = 12.0 V / 2.50 A = 4.8 ohms.

Question 5

Two point charges of +2.0 nC and +8.0 nC are separated in air by a distance of 0.20 m. What is the magnitude of the electrostatic force between them? (Take k = 9.0 x 109 N m2/C2)

  1. A.3.6 x 10^-6 N
  2. B.7.2 x 10^-6 N
  3. C.3.6 x 10^-4 N
  4. D.1.8 x 10^-5 N
Show the answer and why

Answer: A 3.6 x 10^-6 N

Why

By Coulomb's law: F = k (q1 q2) / r2. Substituting the numbers: F = (9.0 x 109 x 2.0 x 10-9 x 8.0 x 10-9) / (0.20)2 = 1.44 x 10-7 / 0.04 = 3.6 x 10-6 N.

Question 6

What is the electric potential produced by a point charge of +1.0 μC at a distance of 1.0 mm from it? (Take k = 9.0 x 109 N m2/C2)

  1. A.9.0 kV
  2. B.90 kV
  3. C.9.0 MV
  4. D.90 MV
Show the answer and why

Answer: C 9.0 MV

Why

Using V = k Q / r: with Q = 1.0 x 10-6 C and r = 1.0 x 10-3 m, we get V = (9.0 x 109 x 1.0 x 10-6) / (1.0 x 10-3) = 9.0 x 106 V = 9.0 MV.

Question 7

Which of the following is true concerning equipotential lines and electric field lines?

  1. A.Equipotential lines can intersect each other.
  2. B.Work is required to move a test charge along an equipotential line.
  3. C.Equipotential lines are always parallel to electric field lines.
  4. D.Equipotential lines are always perpendicular to electric field lines.
Show the answer and why

Answer: D Equipotential lines are always perpendicular to electric field lines.

Why

Equipotential lines are always perpendicular to electric field lines. Since the potential is constant along an equipotential line, no work is required to move a charge along it, and two equipotential surfaces never intersect.

Question 8

A 4.0 ohm resistor and a 12.0 ohm resistor are connected in parallel. This combination is connected in series with a 5.0 ohm resistor and an 8.0 ohm resistor. What is the total equivalent resistance of the circuit?

  1. A.8.0 ohms
  2. B.16.0 ohms
  3. C.24.0 ohms
  4. D.29.0 ohms
Show the answer and why

Answer: B 16.0 ohms

Why

First, find the parallel resistance: Rp = (4.0 x 12.0) / (4.0 + 12.0) = 48 / 16 = 3.0 ohms. Then add the series resistors: Req = 5.0 + 3.0 + 8.0 = 16.0 ohms.

Question 9

A flat square surface of area 2.0 cm by 2.0 cm lies in the xy-plane. An electric field is given by E = (20 j + 50 k) N/C. What is the electric flux through this square surface?

  1. A.0.008 N m^2/C
  2. B.0.02 N m^2/C
  3. C.0.20 N m^2/C
  4. D.2.8 N m^2/C
Show the answer and why

Answer: B 0.02 N m^2/C

Why

Because the surface is in the xy-plane, its area vector points along the z-axis (k-direction): A = (0.02 m x 0.02 m) k = 4 x 10-4 m2 k. The electric flux is Φ = E • A = (50)(4 x 10-4) = 0.02 N m2/C.

Question 10

A galvanometer has a coil resistance of 50 ohms and gives full-scale deflection with a current of 5 mA. What shunt resistance must be connected in parallel to convert it into an ammeter with a full-scale range of 10 A?

  1. A.0.025 ohms
  2. B.0.25 ohms
  3. C.2.5 ohms
  4. D.50.0 ohms
Show the answer and why

Answer: A 0.025 ohms

Why

The required shunt resistance is Rsh = [Ig / (I - Ig)] x Rg. Substituting the given values: Rsh = [0.005 / (10 - 0.005)] x 50 = (0.005 / 9.995) x 50 ≈ 0.025 ohms.

The app holds 12 questions on this unit, marked as you go. These are the first 10.

The textbook pages for unit 6

Unit 6 runs from page 217 to page 286 of the Ministry of Education edition.

Ministry of Education edition, English, 338 pages

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All Grade 11 Physics short notes

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