A total charge Q = 3.6 μC is distributed uniformly over a

quarter circle arc of radius a = 8.8 cm as shown. a)What is λ the linear charge density along the arc? b)What is Ex, the value of the x-component of the

electric field at the origin (x,y) = (0,0) ? x c)What is Ey, the value of the y-component of the

electric field at the origin (x,y) = (0,0) ? y d) How does the magnitude of the electric field at the origin

for the quarter-circle arc you have just calculated comnpare to the

electric field at the origin produced by a point charge Q = 3.6 μC

located a distance a = 8.8 cm from the origin along a

45o line as shown in the figure? o A. The magnitude of

the field from the point charge is less than that from the

quarter-arc of charge. B. The magnitude of the

field from the point charge is equal to that from the quarter-arc

of charge C. The magnitude of the

field from the point charge is greater than that from the

quarter-arc of charge A. The magnitude of

the field from the point charge is less than that from the

quarter-arc of charge. B. The magnitude of the

field from the point charge is equal to that from the quarter-arc

of charge C. The magnitude of the

field from the point charge is greater than that from the

quarter-arc of charge A total charge Q = 3.6 muC is distributed uniformly over a quarter circle arc of radius a = 8.8 cm as shown.d) How does the magnitude of the electric field at the origin for the quarter – circle arc you have just calculated comnpare to the electric field at the origin produced by a point charge Q = 3.6 muC located a distance a = 8.8 cm from the origin along a 45degree line as shown in the figure? A. The magnitude of the field from the point charge is less than that from the quarter – arc of charge. B. The magnitude of the field from the point charge is equal to that from the quarter – arc of charge C. The magnitude of the field from the point charge is greater than that from the quarter – arc of chargea)What is lambda the linear charge density along the arc?b)What is Ex, the value of the x – component of the electric field at the origin (x,y) = (0,0) ?c)What is Ey, the value of the y – component of the electric field at the origin (x,y) = (0,0) ?

General guidance

Concepts and reason

The concepts used to solve this problem are linear charge density, electric field due to arc, and expression for the magnitude of the vector.

Initially, Use the expression of linear charge density to calculate the linear charge density. Later, use the expression of electric field due to arc to calculate the x and y components of electric field. Finally, Use the expression to calculate the magnitude to calculate the magnitude of the electric field and then use the expression of electric field due to a single charge to find the correct option.

Fundamentals

The linear charge density is expressed as follows:

Here, Q is the charge, is the linear charge density, and L is length of the arc.

The length of the arc is expressed as follows:

Here, a is the radius.

The electric field at origin in x direction is expressed as follows:

Here, is the linear charge density, a is the radius of arc, is the angle made by the arc, is the permittivity of free space, and is the electric field in x direction.

The electric field at origin in y direction is expressed as follows:

Here, is the linear charge density, a is the radius of arc, is the angle made by the arc, is the permittivity of free space, and is the electric field in y direction.

The magnitude of electric field is calculated as follows:

Here, is the electric field in x direction, is the electric field in y direction, and is the magnitude of electric field.

The electric field due to a single charge is expressed as follows:

Here, Q is the charge, is the permittivity of free space, and a is the radius of arc.

Step-by-step

Step 1 of 4

(a)

The linear charge density is expressed as follows:

Convert the charge from micro coulombs to coulombs.

Convert the radius from centimeter to meters.

Substitute for in expression .

Substitute for a and for a in expression .

Part a

The linear charge density is .

The linear charge density is directly proportional to the charge and inversely proportional to the length of the arc.

Do not take because the linear charge density is inversely proportional to the length. Take to calculate the linear charge density.

Use the expression of the electric field to calculate the electric field at origin in horizontal direction.

Step 2 of 4

(b)

The electric field at origin in x direction is expressed as follows:

Solve the integration to calculate the expression to calculate the electric field.

Substitute for , for , and for a in expression .

The x- component of the electric field is directed along negative x- direction.

Part b

The value of the x-component of the electric field at the origin is .

From the figure, the horizontal component of the electric field is given by.

Integrate the expression of electric field from to not to .

Use the expression of electric field to calculate the electric field at origin in vertical direction.

Step 3 of 4

(c)

The electric field at origin in y direction is expressed as follows:

Here, is the linear charge density, a is the radius of arc, is the angle made by the arc, is the permittivity of free space, and is the electric field in y direction.

Integrate the expression to find the expression of electric field in vertical direction.

Substitute for , for , and for a in expression .

The y- component of the electric field is directed along negative y- direction.

Part c

The value of the y-component of the electric field at the origin is .

The expression of electric field is calculated using the expression of electric field at origin.

Use the expression to calculate the magnitude of the electric field and electric field due to single charge to find the correct option.

Step 4 of 4

(d)

The magnitude of electric field is calculated as follows:

Here, is the electric field in x direction, is the electric field in y direction, and is the magnitude of electric field.

Substitute for and for in expression .

The electric field due to a single charge is expressed as follows:

Here, Q is the charge, is the permittivity of free space, and a is the radius of arc.

Substitute for , for Q , and for a in expression .

Part d

The magnitude of the field from the point charge is greater than that from the quarter-arc of charge

The electric field due to a point charge is directly proportional to the magnitude of the charge and inversely proportional to the square of the distance.

Answer

Part a

The linear charge density is .

Part b

The value of the x-component of the electric field at the origin is .

Part c

The value of the y-component of the electric field at the origin is .

Part d

The magnitude of the field from the point charge is greater than that from the quarter-arc of charge

Answer only

Part a

The linear charge density is .

Part b

The value of the x-component of the electric field at the origin is .

Part c

The value of the y-component of the electric field at the origin is .

Part d

The magnitude of the field from the point charge is greater than that from the quarter-arc of charge

E 12 cose = 1 4€,a

E 72 sin = 1 Jade o 41€, a

E = E2 +E,

(10C Q=(3.6°C) 1ục ) = 3.6×10°C

10m a =(8.8cm) 1cm = 8.8×10-2m

8.8×10 m

3.6x10C

2 = 3.6x10C 5(8.8×10-?m) = 2.60x10C/m

10=r

ΕΞΙ λcos θ ,, Ο 4πε,α οοο

Ε. = .[sin oli – Α.(sin(s)-sin (0) λ 4πε, (1-0) 4πε,α

2.6x10C/m

9.0×10° N.m

4πες

8.8×10 m

Ε, Απερα

F E =-[9.0×10°°C ox109 N-m? 2.6×10°C/m 8.8×102 m = -2.6x10N/C

E 12 cose = 1 4€,a

λsinθ ,, Ε, Αθ ) 4πεα ο 4πε,α”

κα(() – εν(ω) – Α, =- 4περα

2.6x10C/m

9.0×10° N.m

4πες

8.8×10 m

Ε, Απερα

E, —(9.0×10° °C 0 | 8.8×10²m) = -2.6x106N/C

E = E2 +E,

-2.6×10′ N/C

-2.7x109N/C

E = E2 +E,

E = (–2.6×10^N/C)’ +(-2.6×10^N/C) = 3.68×10°N/C

9.0×10° N.m

4πες

3.6x10C

8.8×10 m

* -(pour = 4.18x10N/C

2.6x10C/m

2.6x10C/m

5.0×10-SC/m

1.0×10 C/m

8.0x10C/m

-2.6×10′ N/C

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