The magnitude and direction of electric field – problems and solutions

1. Calculate the magnitude and direction of the electric field at a point A located at 5 cm from a point charge Q = +10 μC. k = 9 x 10⁹ Nm²C⁻², 1 μC = 10⁻⁶ C)

Known :

Electric charge (Q) = +10 μC = +10 x 10^-6 C

The distance between point A and point charge Q (r_A) = 5 cm = 0.05 m = 5 x 10^-2 m

k = 9 x 10⁹ Nm²C⁻²

Wanted: The magnitude and direction of the electric field at point A

Solution :

The direction of the electric field at point A :

The electric charge is positive hence the direction of the electric field away from the electrical charge and points A.

2. Calculate the magnitude and direction of the electric field at a point P located at 10 cm from a point charge Q = -20 μC. k = 9 x 10⁹ Nm²C⁻², 1 μC = 10⁻⁶ C.

Known :

Electric charge (q) = -20 μC = -20 x 10^-6 C

The distance between point P and electric charge (r_P) = 10 cm = 0.1 m = 1 x 10^-1 m

k = 9 x 10⁹ Nm²C⁻²

Wanted: The magnitude and direction of the electric field at point P

Solution :

The direction of electric field at point A :

The electric charge is negative hence the direction of the electric field to the electrical charge.

3. Two point charges are separated by a distance of 40 cm. What is the magnitude and direction of the electric field at the point P between the two charges, that is 20 cm from point A?

Known :

Charge A (q_A) = -2 μC = -2 x 10^-6 C

Charge B (q_B) = +4 μC = +4 x 10^-6 C

The distance between charge A and point P (r_AP) = 20 cm = 0.2 m = 2 x 10^-1 m

The distance between charge B and point P (r_BP) = 20 cm = 0.2 m = 2 x 10^-1 m

Wanted : The magnitude and direction of electric field at point P.

Solution :

Charge A is negative so that the direction of the electric field points toward Q_A (to the left).

Charge B is positive so that the direction of the electric field points away from Q_B (to the left).

The total electric field at point A :

E = E_A + E_B

E = (4.5 x 10⁵) + (9 x 10⁵)

E = 13.5 x 10⁵ N/C

The direction of the electric field points toward Q_A (to the left).

4. The magnitude of the electric field is zero at…

The charge A is positive and the charge B is positive so that the magnitude of the electric field is zero located at point P, between both charges.

Known :

Charge A (q_A) = +20 μC = +20 x 10⁻⁶ C

Charge B (q_B) = +40 μC = +40 x 10⁻⁶ C

k = 9 x 10⁹ Nm²C⁻²

The distance between charge A and the charge B = 20 cm

The charge between charge A and point P (r_AP) = a

The distance between charge B and point P (r_BP) = 20 – a

Wanted : The magnitude of the electric field is zero located at….

Solution :

The magnitude of the electric field produced by charge A at point P

Charge A is positive so that the direction of the electric field points away from charge A (to the right).

The magnitude of the electric field produced by charge B at point P :

Charge B is positive so that the direction of the electric field points away from charge B (to the left).

The total electric field at point P = 0 :

We use the quadratic formula to determine a.

The magnitude of the electric field is zero located at 8 cm from charge A or 12 cm from charge B.

5. Based on the figure below, where is the point P so that the electric field at point P is zero? (k = 9 x 10⁹ Nm²C⁻², 1 μC = 10⁻⁶ C)

Solution

To calculate the electric field strength at point P, assumed at point P there is a positive test charge. Q₁ is positive and Q₂ is negative, therefore point P must be on the right of Q₂ or left of Q₁. If point P is to the left of Q₁; the electric field generated by Q₁ at the point P is to the left (away from Q₁) and the electric field generated by Q₂ at the point P to the right (towards Q₁). The direction of the electric field is opposite so that the two electric fields eliminate each other so that the electric field strength at point P is zero.

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 : location of point P so that the electric field at point P is zero

Solution :

Point P is on the left of Q₁.

The electric field produced by Q₁ at point P:

The test charge is positive and Q₁ is positive so that the direction of an electric field to leftward.

The electric field produced by Q₂ at point P:

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

Net electric field at point A :

Use the quadratic formula to determine a :

a = -1.25, b = -13.5, c = -20.25

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

Point P is on the 1.2 cm right of Q₁.