Electric Charge and Electric Field Example Problems with Solutions PDF

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Electric Charge and Electric Field: Example Problems with Solutions1. STATIC ELECTRICITY AND CHARGE: CONSERVATION OF CHARGECommon static electricity involves charges ranging from nanocoulombs to microcoulombs.a. How many electrons are needed to form a charge of –2 nC?SolutionAll charged objects in n...

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Electric Charge and Electric Field: Example Problems with Solutions 1. STATIC ELECTRICITY AND CHARGE: CONSERVATION OF CHARGE Common static electricity involves charges ranging from nanocoulombs to microcoulombs. a. How many electrons are needed to form a charge of –2.00 nC? Solution All charged objects in nature carry charges that are integral multiples of the basic quantity of charge, qe, any charge Q: Q = n qe ∣ qe ∣ = 1.60×10−19 C.

N

Q  2.00  10 9 C N N  1.25 10 10 electrons 19 qe  1.60  10 C

b. How many electrons must be removed from a neutral object to leave a net charge of 0.500 µC? N

 0.500  106 C  1.60 10 19 C

N  3.13 10 12 electrons

2. COULOMB’S LAW Three charges, lie along the x axis as shown: q1 = 6 µC, q2 = -2 µC. Determine the magnitude and direction of the net force on q3 = 1.5 µC. q q 2m 1m Fk 12 2 r F1 F2 q =-2 µC q =6 µC q =1.5 µC 2

3

1

Solution There are two Coulomb forces acting on q3:  F1 is the force acting on q3 due to q1.  F2 is the force acting on q3 due to q2. We can find the net force by vectorially adding these two forces. First we need to draw the free-body diagram for q3: we will draw the two forces (vectors) acting on this charged particle. F1 is directed to the right, because q1 and q3 repel each other and F2 is also directed to the right, because q3 and q2 attract each other. We will determine the magnitude of each force considering the absolute values of the charges. Then we will add vectorially the two forces, considering their direction.

F1  k F2  k

q1 q 3 r12 q2 q 3 r22



6 10 C 1.5 10 C N  m /C  6

9

F1  9  10



2

F2  9  109 N  m2 /C2

F net F 1 F 2 0.081 N 0.007 N

6

2

F1  0.081 N

1 m 2

 2  10 C1.5  10 C 6

6

2 m2

1

F 2  0.00675 N  0.007 N

Fnet  0.088 N

3. ADDING ELECTRIC FIELDS Problem Solving Strategy: Electric Fields due to Point Charges a. Electric field depends on the position: choose the point where you want to determine the field. b. Draw a diagram: Draw the electric field vector at that point due to each charge. The direction is given by the direction of the force on a positive test charge. c. Use equation to find the magnitude of the electric field at that particular point due to the individual charges d. The Superposition Principle applies if more than one charge is present

Example 1: Find the electric field at a point P located midway between the charges when both charges are positive as shown. Solution Each point charge creates an electric field of its own at this particular point, therefore there are two electric field vectors acting at point P:  E1 is the electric field at P field due to q1.  E2 is the electric field at P field due to q2.

2m 1m q1 = +2.5 µC

E2 P

E1 q2 = +2.5 µC

We can find the net field by vectorially adding these two vectors. Since the two electric field vectors are equal in magnitude and opposite in direction, they cancel each other out so the net field at P is zero. q q E  k 2 q1  q2 r1  r2  1m E1  E2  k 2 Enet  0 r r Note: An electric field exists in the region of space around a charged object if there is another charged object at that location or not.

2

Example 2: Find the magnitude and direction of the total electric field at the origin of the coordinate system due to the two point charges, q1 and q2. The two charges are located at the x-y coordinate position of (0.0, -2.0 cm) and (+4.00 cm, 0.0), respectively, as shown in the figure. y

Solution We first must find the electric field due to each charge at the point of interest, which is the origin of the coordinate system (O) in this instance. We draw the electric field vectors E1 and E2 on the diagram at the origin of the coordinate system, keeping in mind that electric fields go away from a positive charge and into a negative one.

E1

E2

q1 r12

E2 =k

q2 r22

q1 = +5 nC

r1  Distancefrom q 1 to O. r1  2 cm  2  10 -2 m

E1  9 109 N  m 2 /C 2 

5 10-9 C

2 10

-2

x

4 cm

2cm

E1  k

q2 = +10 nC

O

E1  1.12  105 N/C

m

2

r2 = Distance from q2 to O. r2 =4 cm=4×10-2 m



E2 = 9×109 N×m2 /C2



10×10-9 C

 4×10

-2

m



E 2 =0.56 × 105 N/C

2

The arrows representing the two vectors form a right triangle and the vector sum can be determined using the Pythagorean Theorem: y E  E 21  E22 E

 1.12 10

5

  2

5



N/C  0.56  10 N/C

Etot 2

5

E  1.25 10 N/C

E1

Find the direction : E1 θ  tan E2 -1

θ 5

θ  tan

-1



1.12  10 N/C - 0.56  105 N/C



θ   63.4

0

E2

-63.40 2cm

This is the angle above the negative x axis. we always want to indicate the angle measured counterclockwise from the positive x axis. So: θ = 1800 – 63.40 θ = 116.60

3

q2 = +10 nC

O 4 cm q1 = +5 nC

x

Example 3: Three point charges are located at the corners of an equilateral triangle, as shown. a. Calculate the electric field at a point P located midway between the two charges on the x axis. b. If a charge of 1 nC is placed at P, determine the force (direction and magnitude) acting on this particle? Solution a. Calculate the electric field at a point P located midway between the two charges on the x axis. Each point charge creates an electric field of its own at point P, therefore there are 3 electric field vectors acting at point P:  E1 is the electric field at P due to q1, pointing away from this positive charge.  E2 is the electric field at P field due to q2, also away from q2. q1 =  E3 is the electric field at P due to q3, pointing toward this negative charge. o Draw diagram: o Use equation to find the magnitude of the electric field at that particular point due to the individual charges E1  k

q1 2

r1

r1  Distancefrom q1 to point P.

E2

r1  0.5m sin 60  0.433 m 0

3  10-9 C E 1  144 N/C E1  9 109 N  m 2 /C 2 0.433m2 -9 q2 9 2 2 8 10 C r2  0.250 m E 2  9  10 N  m /C  E2  k 2 r2 0.250 m  2



E3  k

q2 =

E1

E2

q3 =



q2 r

2 3



E 3  9  10 9 N  m 2/C 2

r3  0.250 m

 5 10

E2  1 150 N/C

-9

C

E3  719 N/C

0.250 m  2

o Apply the superposition principle: Add the three vectors. Note that E2 and E3 are both in the positive x direction and E1 is in the negative y direction. q1 = 2 2 3 E

 E2  E3 

 E1

E  1.88 10 N/C

Find the direction : - 144 N/C E1 θ  tan-1 θ  tan-1 E 2  E 3  1869 N/C

θ   4.40

θ  3600  4.40

θ

θ  355.60

b. If a charge of 1 nC is placed at P, determine the force (direction and magnitude) acting on this particle? E

F  F q E q





F  1.00 10 -9 C 1.88 10 3 N/C

 S ame direction as E : θ  355.6 0

4



q2 = F  1.88 10 -6 N

Ey q 3 =

Ex

E...