Electromagnetic Waves, Antennas, and RF: Electric Charge and the Electric Field - Exercises and Complete Worked Solutions

This companion article provides self-study exercises for EM05, Electric Charge and the electric field. All exercises are stated first. Complete worked solutions follow in Part II.

The problems reinforce the central ideas introduced in EM05: charge sign and quantization, Coulomb’s law, source-to-observation geometry, the test-charge definition of electric field, the field of a point charge, inverse-square scaling, and vector superposition.

The elementary charge magnitude is

(1)

Coulomb’s law for two stationary point charges is

(2)

The electric field is defined by

(3)

or equivalently

(4)

For a point charge q at source point r′, the field at observation point r is

(5)

Electric fields superpose linearly:

(6)

These are the same definitions and conventions used in EM05 [1, 2, 3, 5].

How to use this problem set

Attempt all exercises in Part I before reading Part II. In vector problems, keep the following sequence explicit:

1. identify source and observation locations;
2. form R = r − r′;
3. determine the magnitude R and unit vector R;
4. apply the sign of the source charge correctly;
5. add field vectors component by component.

Part I: Exercises

Exercise 1: charge quantization

An isolated object has net charge

(7)

How many excess electrons does this charge represent? State the sign interpretation clearly.

Exercise 2: Coulomb-force magnitude and sign

Two point charges are separated by

(8)

Their charges are

(9)

Find the magnitude of the electrostatic force and state whether the interaction is attractive or repulsive.

Exercise 3: vector Coulomb force from coordinates

A source charge

(10)

is located at

(11)

and a second charge

(12)

is located at

(13)

Find the force on q₂ due to q₁ in Cartesian-vector form.

Exercise 4: source point, observation point, and separation vector

A point charge is located at

(14)

The observation point is

(15)

Find:

1. R = r − r′;
2. R = |R|;
3. R.

Exercise 5: field magnitude and direction from a negative point charge

A point charge

(16)

is at the origin. Find the electric-field magnitude at a distance

(17)

and state the field direction relative to the source charge.

Exercise 6: force on a negative test charge

At a point in space,

(18)

A test charge

(19)

is placed at that point.

Find the force vector and its magnitude. Explain why the force direction is opposite the electric-field direction.

Exercise 7: cancellation by symmetry

Two identical positive charges +q are placed at

(20)

Find the total electric field at the origin. Explain the cancellation using both magnitude and direction.

Exercise 8: two-dimensional vector superposition

At an observation point P, two sources produce

(21)

and a second field of magnitude

(22)

at an angle of 120^∘ counterclockwise from the +x axis.

Find the total field in Cartesian form, its magnitude, and its direction measured counterclockwise from +x.

Exercise 9: opposite charges and a midpoint field

A charge

(23)

is located at x = −0.20 m, and a charge

(24)

is located at x = +0.20 m.

Find the total electric field at the origin. State its direction.

Exercise 10: inverse-square scaling

A point charge produces field magnitude E₁ at distance r₁.

Find E₂∕E₁ when

(25)

Then find the distance ratio r₂∕r₁ required to reduce the field magnitude to one ninth of its original value.

Exercise 11: field versus force

A uniform electric field at a point is

(26)

Find the force on each of the following particles:

1. a proton, charge +e;
2. an electron, charge −e.

Explain why the same field produces opposite force directions.

Exercise 12: write the continuous-source field integral

A volume charge density ρ(r′) occupies a finite source region V ′. Using source coordinate r′ and observation coordinate r, write the integral expression for the electric field at r.

Then identify the physical role of each of the following:

(27)

No integration is required.

Exercise 13: diagnose conceptual statements

For each statement, decide whether it is correct. If it is incorrect, rewrite it accurately.

1. “A negative source charge produces a negative electric-field magnitude.”
2. “If the electric field at a point is zero, every individual source field there must also be zero.”
3. “The vector r′ labels the source point and is not a derivative.”
4. “A negative test charge experiences force opposite to the electric field.”
5. “The point-charge field falls as 1∕r.”

Exercise 14: synthesis - source geometry, field, and force

A source charge

(28)

is located at

(29)

The observation point is

(30)

A test charge

(31)

is placed at the observation point.

Find:

1. R and R;
2. the electric field vector E(r);
3. the field magnitude;
4. the force vector on the test charge;
5. the force magnitude;
6. whether the force points toward or away from the source charge.

Part II: Complete Worked Solutions

Solution 1: charge quantization

Charge quantization gives

(32)

Therefore,

N	=	(33)
	=	(34)
	≈−3.	(35)

Thus the object has three excess electrons:

(36)

The negative sign in N = −3 indicates excess negative elementary charges.

Solution 2: Coulomb-force magnitude and sign

The force magnitude is

(37)

Substituting,

F	= (8.99 × 109)	(38)
	≈ 3.37 × 10−7 N.	(39)

Therefore,

(40)

Because the charges have opposite signs, the force is attractive.

Solution 3: vector Coulomb force from coordinates

The vector from q₁ to q₂ is

R	= r2 − r1	(41)
	= (4 − 1)x + (6 − 2)y	(42)
	= 3x + 4y m.	(43)

Its magnitude is

(44)

The unit vector is

(45)

The force magnitude is

F	= ke	(46)
	= (8.99 × 109)	(47)
	≈ 3.60 × 10−9 N.	(48)

Both charges are positive, so the force on q₂ points away from q₁, in the +R direction. Hence

F2←1	= FR	(49)
	≈ (3.60 × 10−9) N.	(50)

Thus

(51)

Solution 4: source point, observation point, and separation vector

The separation vector is

R	= r − r′	(52)
	= (5 − 2)x + [3 − (−1)]y + (3 − 3)z	(53)
	= 3x + 4y m.	(54)

Therefore,

(55)

Its magnitude is

(56)

The unit vector is

(57)

Solution 5: field magnitude and direction from a negative point charge

The field magnitude is

E	= ke	(58)
	= (8.99 × 109)	(59)
	≈ 5.75 × 102 N/C.	(60)

Thus

(61)

Because the source charge is negative, the field points radially toward the charge.

Solution 6: force on a negative test charge

Use

(62)

Then

F	= (−6.0 × 10−9)(250x − 150y) N	(63)
	= (−1.50 × 10−6x + 9.00 × 10−7y) N.	(64)

Therefore,

(65)

Its magnitude is

F	= μN	(66)
	≈ 1.75 μN.	(67)

Hence

(68)

The negative sign of q_t reverses the force direction relative to E.

Solution 7: cancellation by symmetry

Each charge is a distance a from the origin, so the two field magnitudes are equal:

(69)

The field from the left positive charge points toward +x, while the field from the right positive charge points toward −x. Therefore,

(70)

Hence

(71)

The cancellation occurs because the fields have equal magnitude and opposite direction.

Solution 8: two-dimensional vector superposition

The first field is

(72)

The second field has components

E2x	= 300 cos 120∘ = −150 N/C,	(73)
E2y	= 300 sin 120∘ ≈ 259.8 N/C.	(74)

Thus

(75)

Add the vectors:

Etotal	= E1 + E2	(76)
	= (250x + 259.8y) N/C.	(77)

Therefore,

(78)

The magnitude is

Etotal	= N/C	(79)
	≈ 360.6 N/C.	(80)

The direction is

𝜃	= tan −1	(81)
	≈ 46.1∘.	(82)

Hence

(83)

Solution 9: opposite charges and a midpoint field

At the origin, each source is 0.20 m away. The magnitude from either charge is

Eeach	= ke	(84)
	≈ 674 N/C.	(85)

The positive charge at x = −0.20 m produces a field toward +x at the origin. The negative charge at x = +0.20 m also produces a field toward +x because the field points toward a negative charge.

Therefore the two fields add:

(86)

Thus

(87)

Solution 10: inverse-square scaling

For a point charge,

(88)

Therefore,

	= 2	(89)
	= 2	(90)
	= .	(91)

So

(92)

For the second part, require

(93)

Then

(94)

so

(95)

Solution 11: field versus force

For a proton,

(96)

Therefore,

Fp	= qpE	(97)
	= (1.602 × 10−19)(500)y N	(98)
	≈ 8.01 × 10−17y N.	(99)

Thus

(100)

For an electron,

(101)

Hence

(102)

The field is the same in both cases. The force reverses because the particle charges have opposite signs.

Solution 12: write the continuous-source field integral

The electric field is

(103)

The factors have the following meanings:

• ρ(r′) is the charge density at the source point;
• dV ′ is an infinitesimal source volume;
• r − r′ is the vector from the source element to the observation point;
• |r − r′|³ combines with the separation vector so that the magnitude scales as 1∕R² while retaining direction.

The prime distinguishes the integration/source coordinate from the observation coordinate.

Solution 13: diagnose conceptual statements

1. Incorrect. Electric-field magnitude is nonnegative. A negative source charge reverses the field direction so that the field points toward the source.
2. Incorrect. Nonzero individual source fields can cancel vectorially, producing zero total field.
3. Correct. The prime is a source-coordinate label, not differentiation.
4. Correct. Since F = q_tE, a negative test charge reverses the force direction relative to E.
5. Incorrect. The electrostatic point-charge field magnitude decreases as 1∕r².

Solution 14: synthesis - source geometry, field, and force

The separation vector is

R	= r − r′	(104)
	= (4 − 1)x + [2 − (−2)]y	(105)
	= 3x + 4y m.	(106)

Thus

(107)

The point-charge field is

(108)

Substituting,

E	= (8.99 × 109)(−8.0 × 10−9) N/C	(109)
	≈ (−1.73x − 2.30y) N/C.	(110)

Therefore,

(111)

The field magnitude is

(112)

The force on the positive test charge is

F	= qtE	(113)
	= (5.0 × 10−9)(−1.73x − 2.30y) N	(114)
	≈ (−8.63x − 11.5y) × 10−9 N.	(115)

Thus

(116)

Its magnitude is

(117)

Because the source charge is negative and the test charge is positive, the force points toward the source charge.

Common mistakes

• Forgetting charge signs. Use magnitudes for scalar force size, but use signed charge in vector formulas.
• Using the wrong separation vector. For a source at r′ and observation point r, use R = r − r′.
• Adding field magnitudes instead of vectors. Superposition is componentwise vector addition.
• Confusing field direction with force direction. A negative test charge experiences force opposite to E.
• Using 1∕r instead of 1∕r². The electrostatic point-charge field is inverse square.
• Treating r′ as a derivative. The prime is only a source-coordinate label.

What EM05E reinforces

The essential field relation is

(118)

It combines source strength, source-to-observation geometry, inverse-square magnitude, and vector direction in one expression.

For many sources,

(119)

and for a continuous charge distribution the sum becomes an integral over source points.

These ideas prepare the next article, EM06, where electric flux and oriented area vectors will be introduced before Gauss’s Law.

References