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 magnitude and direction of electric field - problems and solutions 1

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 magnitude and direction of electric field - problems and solutions 2

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?

The magnitude and direction of electric field - problems and solutions 3

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 :

The magnitude and direction of electric field - problems and solutions 4

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

The magnitude and direction of electric field - problems and solutions 5

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 magnitude and direction of electric field - problems and solutions 6

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

The magnitude and direction of electric field - problems and solutions 7

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 :

The magnitude and direction of electric field - problems and solutions 8

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 :

The magnitude and direction of electric field - problems and solutions 9

We use the quadratic formula to determine a.

The magnitude and direction of electric field - problems and solutions 10

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)

The magnitude and direction of electric field - problems and solutions 11

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 magnitude and direction of electric field - problems and solutions 12

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 magnitude and direction of electric field - problems and solutions 13

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 :

The magnitude and direction of electric field - problems and solutions 14

Use the quadratic formula to determine a :

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

The magnitude and direction of electric field - problems and solutions 15

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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