# Electric field – problems and solutions

1. Point A located at the center between two charges. Both charges have the same magnitude but opposite sign and separated by a distance of a. The magnitude of the electric field at point A is 36 N/C. If point A moved 1/2a close to one of both charges, what is the magnitude of the electric field at point A?

Known :

Charge 1 (q1) = +Q

Charge 2 (q2) = -Q

The distance between charge 1 and point A (r1A) = ½ a

The distance between charge 2 and point A (r2A) = ½ a

The magnitude of the electric field at point A (EA) = 36 NC-1

Wanted: The magnitude of the electric field

Solution :

Step 1.

The electric field produced by a charge +Q at point A : Test charge is positive and charges 1 is positive so that the direction of the electric field points to charge 2.

The electric charge produced by a charge -Q at point A : Test charge is positive and charges 2 is negative so that the direction of the electric field points to charge 2.

The resultant of the electric field at point A : Step 2.

If point A is moved close to charge 1 then :

The distance between charge 1 and point A (r1A) = ¼ a

The distance between charge 2 and point A (r2A) = ¾ a

The electric field produced by charge +Q at point A : Test charge is positive and charges 1 is positive so that the direction of the electric field points to charge 2.

The electric field produced by charge -Q at point A : Test charge is positive and charges 2 is negative so that the direction of the electric field points to charge 2.

The resultant of the electric field at point A : Read :  Motion on the horizontal surface without the friction force – application of Newton's law of motion problems and solutions

2. Two charges qA = 1 μC and qB = 4 μC are separated by a distance of 4 cm (k = 9 x 109 N m2 C−2). What is the magnitude of the electric field at the center between qA and qB.

Known :

Charge A (qA) = 1 μC = 1 x 10−6 C

Charge B (qB) = 4 μC = 4 x 10−6 C

k = 9 x 109 N m2 C−2

Distance between charge A and B (rAB) = 4 cm = 0.04 meters

Distance between charge A and the center point (rA) = 0.02 meters

Distance between charge B and the center point (rB) = 0.02 meters

Known: The magnitude of the electric field

Solution :

The electric field produced by charge A at the center point : Test charge is positive and charges A is positive so that the direction of the electric field points to charge B.

The electric field produced by charge B at the center point : Test charge is positive and charge B is positive so that the direction of the electric field points to charge A.

The resultant of the electric field at the center point :

EA and EB have the opposite direction.

E = EB – EA = 9 x 107 – 2.25 x 107 = 6.75 x 107 NC-1

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3. According to figure below, where the point P is located so that the magnitude of the electric field at point P = 0 ? (k = 9 x 109 Nm2C−2, 1 μC = 10−6 C) Solution

If point P located at the left of Q1; the electric field produced by Q1 on point P points to leftward (away from Q1) and the electric field produced by Q2 on point P points to rightward (point to Q1). The direction of the electric field is opposite so that the electric field at point P = 0.

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 : Position of point P

Solution :

Point P located at leftward of Q1.

The electric field produced by Q1 at point P : Test charge is positive and Q1 is positive so that the direction of the electric field to leftward.

The electric field produced by Q2 at point P : Test charge is positive and Q2 is negative so that the direction of the electric field to rightward.

Resultant of the electric field at point A : Use quadratic formula to find a : Distance between Q2 and point P (r2P) = 3 + a = 3 – 1.8 = 1.2 cm or 3 + a = 3 – 9 = -6 cm.

Distance between Q1 and point P (r1P) = a = -9 cm or -1.8 cm.

Point P located at 1.2 cm rightward of Q2.

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4. Charge q3 located at 5 cm rightward of q2, as shown in the figure below. What is the magnitude of the electric field at charge q3 (1 µC = 10-6 C). Solution : Charge q3 is positive so that the direction of the electric field at charge q3 points to the minus charge q2 (E2) and away from the plus charge q1 (E1). The resultant of the electric field is the sum of the electric field E1 and E2.

Known :

Charge q1 = 5 µC = 5 x 10-6 Coulomb

Charge q2 = 5 µC = -5 x 10-6 Coulomb

Distance between charge q1 and charge q3 (r1) = 15 cm = 0.15 m = 15 x 10-2 meters

Distance between charge q2 and charge q3 (r2) = 5 cm = 0.05 m = 5 x 10-2 meters

k = 9 x 109 N m2 C-2

Wanted : The electric field at charge q3

Solution :

The electric field 1 :

E1 = k q1 / r12

E1 = (9 x 109)(5 x 10-6) / (15 x 10-2)2

E1 = (45 x 103) / (225 x 10-4)

E1 = 0.2 x 107 N/C

The electric field 2 :

E2 = k q2 / r22

E2 = (9 x 109)(5 x 10-6) / (5 x 10-2)2

E2 = (45 x 103) / (25 x 10-4)

E2 = 1.8 x 107 N/C

Resultant of the electric field :

The resultant of the electric field at charge q3 :

E = E2 – E1 = (1.8 x 107) – (0.2 x 107) = 1.6 x 107 N/C

The direction of the electric field points to leftward (same direction as E2).

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5. Two charges are separated as shown in figure below. What is the electric field at point P (k = 9 x 109

N m2 C-2)

Solution Known :

Charge qA = +2.5 C

Charge qB = -2 C

Distance between charge qA and point P (rA) = 5 m

Distance between charge qB and point P (rB) = 2 m

k = 9 x 109 N m2 C-2

Wanted : the magnitude of the electric field at point P.

Solution :

The electric field A :

EA = k qA / rA2

EA = (9 x 109)(2.5) / (5)2

EA = (22.5 x 109) / 25

EA = 0.9 x 109 N/C

The electric field B :

EB = k qB / rB2

EB = (9 x 109)(2) / (2)2

EB = (18 x 109) / 4

EB = 4.5 x 109 N/C

Resultant of the electric field :

Resultant of the electric field at point P :

E = EB – EA = (4.5 – 0.9) x 109 = 3.6 x 109 N/C

The direction to leftward (same direction as EB).

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6. Two charges Q1 = -40 µC and Q2 = +5 µC as shown in figure below (k = 9 x 109 N.m2.C-2 and 1 µC = 10-6 C),. What is the magnitude of the electric field at point P. Known :

Charge q1 = -40 µC = -40 x 10-6 C

Charge q2 = +5 µC = +5 x 10-6 C

Distance between q1 and point P (r1) = 40 cm = 0.4 m = 4 x 10-1 m

Distance between q2 and point P (r2) = 10 cm = 0.1 = 1 x 10-1 m

k = 9 x 109 N m2 C-2

Wanted : the magnitude of the electric field at point P.

Solution :

The electric field 1 :

E1 = k q1 / r12

E1 = (9 x 109)(40 x 10-6) / (4 x 10-1)2

E1 = (360 x 103) / (16 x 10-2)

E1 = 22.5 x 105 N/C

The electric field 2 :

E2 = k q2 / r22

E2 = (9 x 109)(5 x 10-6) / (1 x 10-1)2

E2 = (45 x 103) / 1 x 10-2

E2 = 45 x 105 N/C

Resultant of the electric field :

The resultant of the electric field at point P :

E = E2 – E1 = (45 – 22.5) x 105 = 22.5 x 105 N/C

E = 2.25 x 106 N/C

The direction of the electric field points to rightward (same direction as E2).

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7. Two point charges as shown in figure below. Where is point P located so that the magnitude of the electric field at point P = 0. k = 9.109 Nm2.C-2, 1 µC = 10-6 C.

Known :

Charge 1 (q1) = -9 µC = -9.10-6 Coulomb

Charge 2 (q2) = 1 µC = 1.10-6 Coulomb

Distance between q1 and q2 (r12) = 1 cm

k = 9.109 Nm2.C-2

Wanted : Position of point P

Solution :

E1 = the magnitude of the electric field produced by q1 at point P

The direction of E1 to q1 because q1 is negative.

E2 = the magnitude of the electric field produced by q2 at point P

The direction of E2 away from q2 because q2 is positive.  The electric field at point = 0. Distance between P and q2 = x = 0.5 cm.

Point P located at 0.5 cm rightward q2 or 0.25 cm leftward q1.

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8. According to the figure below, if the magnitude of the electric field at point P = k Q/x2, then x = …. Known :

EP = k Q / x2

Wanted : x

Solution : E2 = The magnitude of the electric field at point P by charge +32Q

r2 =Distance between charge +32Q and point P = a + x  