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.

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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 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.

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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 (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).

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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 (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.

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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 10^{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}.

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