# 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 10^{9 }Nm^{2}C^{−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 10^{9 }Nm^{2}C^{−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 10^{9 }Nm^{2}C^{−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 10^{9 }Nm^{2}C^{−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 10^{5}) + (9 x 10^{5})

E = 13.5 x 10^{5 }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

__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 10^{9} Nm^{2}C^{−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?

^{9 }Nm

^{2}C

^{−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. Q_{1} is positive and Q_{2} is negative, therefore point P must be on the right of Q_{2 }or left of Q_{1}. If point P is to the left of Q_{1}; the electric field generated by Q_{1 }at the point P is to the left (away from Q_{1}) and the electric field generated by Q_{2 }at the point P to the right (towards Q_{1}). 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 :__

Q_{1} = +9 μC = +9 x 10^{−6} C

Q_{2} = -4 μC = -4 x 10^{−6} C

k = 9 x 10^{9 }Nm^{2}C^{−2}

Distance between charge 1 and charge 2 = 3 cm

Distance between Q_{1} and point P (r_{1P}) = a

Distance between Q_{2} 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 Q_{1}.

The electric field produced by Q_{1} at point P:

The test charge is positive and Q_{1} is positive so that the direction of an electric field to leftward.

The electric field produced by Q_{2} at point P:

The test charge is positive and Q_{2} 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 Q_{2} and point P (r_{2P}) = 3 + a = 3 – 1.8 = 1.2 cm.

Point P is on the 1.2 cm right of Q_{1}.