Electric field – problems and solutions

1. Point A located at the center between two charges. Both charges have the same magnitude but opposite sign and separated by a distance of a. The magnitude of the electric field at point A is 36 N/C. If point A moved 1/2a close to one of both charges, what is the magnitude of the electric field at point A?

Known :

Charge 1 (q₁) = +Q

Charge 2 (q₂) = -Q

The distance between charge 1 and point A (r_1A) = ½ a

The distance between charge 2 and point A (r_2A) = ½ a

The magnitude of the electric field at point A (E_A) = 36 NC^-1

Wanted: The magnitude of the electric field

Solution :

Step 1.

The electric field produced by a charge +Q at point A :

Test charge is positive and charges 1 is positive so that the direction of the electric field points to charge 2.

The electric charge produced by a charge -Q at point A :

Test charge is positive and charges 2 is negative so that the direction of the electric field points to charge 2.

The resultant of the electric field at point A :

Step 2.

If point A is moved close to charge 1 then :

The distance between charge 1 and point A (r_1A) = ¼ a

The distance between charge 2 and point A (r_2A) = ¾ a

The electric field produced by charge +Q at point A :

Test charge is positive and charges 1 is positive so that the direction of the electric field points to charge 2.

The electric field produced by charge -Q at point A :

Test charge is positive and charges 2 is negative so that the direction of the electric field points to charge 2.

The resultant of the electric field at point A :

2. Two charges q_A = 1 μC and q_B = 4 μC are separated by a distance of 4 cm (k = 9 x 10⁹ N m² C⁻²). What is the magnitude of the electric field at the center between q_A and q_B.

Known :

Charge A (q_A) = 1 μC = 1 x 10⁻⁶ C

Charge B (q_B) = 4 μC = 4 x 10⁻⁶ C

k = 9 x 10⁹ N m² C⁻²

Distance between charge A and B (r_AB) = 4 cm = 0.04 meters

Distance between charge A and the center point (r_A) = 0.02 meters

Distance between charge B and the center point (r_B) = 0.02 meters

Known: The magnitude of the electric field

Solution :

The electric field produced by charge A at the center point :

Test charge is positive and charges A is positive so that the direction of the electric field points to charge B.

The electric field produced by charge B at the center point :

Test charge is positive and charge B is positive so that the direction of the electric field points to charge A.

The resultant of the electric field at the center point :

E_A and E_B have the opposite direction.

E = E_B – E_A = 9 x 10^{7 –} 2.25 x 10⁷ = 6.75 x 10⁷ NC^-1

3. According to figure below, where the point P is located so that the magnitude of the electric field at point P = 0 ? (k = 9 x 10⁹ Nm²C⁻², 1 μC = 10⁻⁶ C)

Solution

If point P located at the left of Q₁; the electric field produced by Q₁ on point P points to leftward (away from Q₁) and the electric field produced by Q₂ on point P points to rightward (point to Q₁). The direction of the electric field is opposite so that the electric field at point P = 0.

Known :

Q₁ = +9 μC = +9 x 10⁻⁶ C

Q₂ = -4 μC = -4 x 10⁻⁶ C

k = 9 x 10⁹ Nm²C⁻²

Distance between charge 1 and charge 2 = 3 cm

Distance between Q₁ and point P (r_1P) = a

Distance between Q₂ and point P (r_2P) = 3 + a

Wanted : Position of point P

Solution :

Point P located at leftward of Q₁.

The electric field produced by Q₁ at point P :

Test charge is positive and Q₁ is positive so that the direction of the electric field to leftward.

The electric field produced by Q₂ at point P :

Test charge is positive and Q₂ is negative so that the direction of the electric field to rightward.

Resultant of the electric field at point A :

Use quadratic formula to find a :

Distance between Q₂ and point P (r_2P) = 3 + a = 3 – 1.8 = 1.2 cm or 3 + a = 3 – 9 = -6 cm.

Distance between Q₁ and point P (r_1P) = a = -9 cm or -1.8 cm.

Point P located at 1.2 cm rightward of Q₂.

4. Charge q₃ located at 5 cm rightward of q₂, as shown in the figure below. What is the magnitude of the electric field at charge q₃ (1 µC = 10^-6 C).

Solution :

Charge q₃ is positive so that the direction of the electric field at charge q₃ points to the minus charge q₂ (E₂) and away from the plus charge q₁ (E₁). The resultant of the electric field is the sum of the electric field E₁ and E₂.

Known :

Charge q₁ = 5 µC = 5 x 10^-6 Coulomb

Charge q₂ = 5 µC = -5 x 10^-6 Coulomb

Distance between charge q₁ and charge q₃ (r₁) = 15 cm = 0.15 m = 15 x 10^-2 meters

Distance between charge q₂ and charge q₃ (r₂) = 5 cm = 0.05 m = 5 x 10^-2 meters

k = 9 x 10⁹ N m² C^-2

Wanted : The electric field at charge q₃

Solution :

The electric field 1 :

E₁ = k q₁ / r₁²

E₁ = (9 x 10⁹)(5 x 10^-6) / (15 x 10^-2)²

E₁ = (45 x 10³) / (225 x 10^-4)

E₁ = 0.2 x 10⁷ N/C

The electric field 2 :

E₂ = k q₂ / r₂²

E₂ = (9 x 10⁹)(5 x 10^-6) / (5 x 10^-2)²

E₂ = (45 x 10³) / (25 x 10^-4)

E₂ = 1.8 x 10⁷ N/C

Resultant of the electric field :

The resultant of the electric field at charge q₃ :

E = E₂ – E₁ = (1.8 x 10⁷) – (0.2 x 10⁷) = 1.6 x 10⁷ N/C

The direction of the electric field points to leftward (same direction as E₂).

5. Two charges are separated as shown in figure below. What is the electric field at point P (k = 9 x 10⁹ N m² C^-2)

Solution

Known :

Charge q_A = +2.5 C

Charge q_B = -2 C

Distance between charge q_A and point P (r_A) = 5 m

Distance between charge q_B and point P (r_B) = 2 m

k = 9 x 10⁹ N m² C^-2

Wanted : the magnitude of the electric field at point P.

Solution :

The electric field A :

E_A = k q_A / r_A²

E_A = (9 x 10⁹)(2.5) / (5)²

E_A = (22.5 x 10⁹) / 25

E_A = 0.9 x 10⁹ N/C

The electric field B :

E_B = k q_B / r_B²

E_B = (9 x 10⁹)(2) / (2)²

E_B = (18 x 10⁹) / 4

E_B = 4.5 x 10⁹ N/C

Resultant of the electric field :

Resultant of the electric field at point P :

E = E_B – EA = (4.5 – 0.9) x 10⁹ = 3.6 x 10⁹ N/C

The direction to leftward (same direction as E_B).

6. Two charges Q₁ = -40 µC and Q₂ = +5 µC as shown in figure below (k = 9 x 10⁹ N.m².C^-2 and 1 µC = 10^-6 C),. What is the magnitude of the electric field at point P.

Known :

Charge q₁ = -40 µC = -40 x 10^-6 C

Charge q₂ = +5 µC = +5 x 10^-6 C

Distance between q₁ and point P (r₁) = 40 cm = 0.4 m = 4 x 10^-1 m

Distance between q₂ and point P (r₂) = 10 cm = 0.1 = 1 x 10^-1 m

k = 9 x 10⁹ N m² C^-2

Wanted : the magnitude of the electric field at point P.

Solution :

The electric field 1 :

E₁ = k q₁ / r₁²

E₁ = (9 x 10⁹)(40 x 10^-6) / (4 x 10^-1)²

E₁ = (360 x 10³) / (16 x 10^-2)

E₁ = 22.5 x 10⁵ N/C

The electric field 2 :

E₂ = k q₂ / r₂²

E₂ = (9 x 10⁹)(5 x 10^-6) / (1 x 10^-1)²

E₂ = (45 x 10³) / 1 x 10^-2

E₂ = 45 x 10⁵ N/C

Resultant of the electric field :

The resultant of the electric field at point P :

E = E₂ – E1 = (45 – 22.5) x 10⁵ = 22.5 x 10⁵ N/C

E = 2.25 x 10⁶ N/C

The direction of the electric field points to rightward (same direction as E₂).

7. Two point charges as shown in figure below.

Where is point P located so that the magnitude of the electric field at point P = 0. k = 9.10⁹ Nm².C^-2, 1 µC = 10^-6 C.

Known :

Charge 1 (q₁) = -9 µC = -9.10^-6 Coulomb

Charge 2 (q₂) = 1 µC = 1.10^-6 Coulomb

Distance between q₁ and q₂ (r₁₂) = 1 cm

k = 9.10⁹ Nm².C^-2

Wanted : Position of point P

Solution :

E₁ = the magnitude of the electric field produced by q₁ at point P

The direction of E₁ to q₁ because q₁ is negative.

E₂ = the magnitude of the electric field produced by q₂ at point P

The direction of E₂ away from q₂ because q₂ is positive.

The electric field at point = 0.

Use quadratic formula :

Distance between P and q₂ = x = 0.5 cm.

Point P located at 0.5 cm rightward q₂ or 0.25 cm leftward q₁.

8. According to the figure below, if the magnitude of the electric field at point P = k Q/x², then x = ….

Known :

E_P = k Q / x²

Wanted : x

Solution :

E₂ = The magnitude of the electric field at point P by charge +32Q

r₂ =Distance between charge +32Q and point P = a + x

Use quadratic formula :