# 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 (rA) = 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.

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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 (rP) = 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.

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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 (qA) = -2 μC = -2 x 10-6 C

Charge B (qB) = +4 μC = +4 x 10-6 C

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

The distance between charge B and point P (rBP) = 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 QA (to the left). Charge B is positive so that the direction of the electric field points away from QB (to the left).

The total electric field at point A :

E = EA + EB

E = (4.5 x 105) + (9 x 105)

E = 13.5 x 105 N/C

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

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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 (qA) = +20 μC = +20 x 10−6 C

Charge B (qB) = +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 (rAP) = a

The distance between charge B and point P (rBP) = 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.

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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 (r1P) = a

Distance between Q2 and point P (r2P) = 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 (r2P) = 3 + a = 3 – 1.8 = 1.2 cm.

Point P is on the 1.2 cm right of Q1.

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