# Electrical Force Test Solutions

## Calculation of force that two or more point electric charges apply to each other. Relationship between electrical forces. Coulomb forces. Solutions of the electrical force questions.

Solution – 1

There are 6.242x1018 electron in electric charge of 1 C. According to this, there are q electron in electric charge of 5 C.

Let we establish the ratio.

1 C charge            6.242x1018 e

5 C charge               q

q = 5*6.242x1018

q = 31.21x1018

The right answer is option B.

Solution – 2

The charge of one electron or proton is 1.6x10-19 C.

The charges of 10 electron is

q = 10*1.6x10-19

q = 1.6x10(-19 + 1)

q = 1.6x10-18 C

Or

q = (10*1.6)x10-18 C

q = 16x10-19 C

The right answer is option C

Solution – 3

Let we find the force F.

 F = k* q1*q2 d2

 F = k* (-6q)*(12q) d2

 F = k* q2 .( -72) d2

Let we do to 3d/5 the distance between the electrical charges. In this case, suppose that F2  is force which the electrical charges apply to each other.

Let we find the magnitude of the force F2.

F2 = k*(-6q)*(12q)
 9d2 25

F2 = ( -72)* k*q2
 * 25 9
d2

 F2 = kq2 (-8)*25 d2

F2
= (-8)kq2 * 25
 * d2 ( -72)* k*q2
d2
F

F2
 = 25*8 72
F

 F2 = 25 F 9

The right answer is option E

Solution – 4

Let's calculate the forces one by one.

 F1 = k* q1*q2 d2

 F1 = k*12 d2

 F1 = k *(12) d2

 F2 = k*18 4d2

F2 = k
 * 9 2
d2

F3 =  k*24
 9d2 4

F3 = k
 * 32 3
d2

All in equations the part of k/d2 are equal. We can suppose this part is 1 N.

F1 = 1*12 N = 12 N

F2 = 1*4.5 N = 4.5 N

F3 = 1*10.7 N = 10.7 N

F1 > F3 > F2

The right answer is option A

Solution – 5

Suppose that the F1 is force applied to the charge q2 by the charge q1 and the F2 is force applied to the charge q2 by the charge q3.

q1 = 6x10-5 C, q2 = 4x10-4 C ve q3 = 5x10-5 C

 F1 = 9x109 * 6x10-5 * 4x10-4 9

 F1 = 9*6*4x10(9 – 5 – 4 9

F1 = 24 N

 F3 = 9x109 * 4x10-4 * 5x10-5 + 4

F3 = 45 N

The both forces attract the q2 charge towards itself.

The resultant force is equal to the absolute value of the F1 – F2.

Whichever force is greater, the direction of the resultant force is in the direction of that force.

F = |45 – 24|

F = 21 N

The magnitude of the resultant force is 21 N and direction of this force is in the direction of the charge q3.

The right answer is option D

Solution – 6

Let we calculate the force that the spheres applies to each other before its not touch.

 F = k* (-18)*(+6) 0.36

F = 300*k

When the spheres touch to the each other, firstly the charge of -6q of the sphere X is neutralize the sphere Y. Then the charge of the sphere X to be -12q. These electrical charges are shared between two spheres. The last charge of the spheres to be -6q. In this case, let’s calculate the force that the spheres applies to each other.

 F2 = k* (-6)*(-6) 0.36

 F2 = k* 36 0.36

F2 = 100*k

F
 = 300 100
F2

 F2 = F 3

The right answer is option B

Solution – 7

 F = k* q1*q2 d2

 F = 9x109 * (1.5)x10-4 *q2 0.25

 324 = 9*(1.5)(q2)x10(9 – 4) 0.25

 36 = (1.5) * (q2)x105 0.25

9 = (1.5) * (q2)x105

 q2 = 9 1.5*105

q2 = 6x10-5 C

The right answer is option E

Solution – 8

 F = k* q1*q2 d2

 43.2 = 9x109 * 6x10-4*8x10-4 d2

 43.2 = 9*6*8x10(9 – 4 – 4 ) d2

 43.2 = 432x10 d2

d2 = 100

d = 10 m

The right answer is option C

Solution – 9

Let we find the force that the charges q1 and q2 applies to each other.

 F = 9x109 5x10-5 * 12x10-5 1

F = 60*9*10-1

F = 54 N

The force F is repel force that the bodies apply to each other. The T is tension force. The mg is weight of the body. The relationship between three magnitudes can found using trigonometric correlations.

 F = sin37° T

 54 = 0.6 T

T = 90 N

If we want to find the mg, we can find it as the follows.

 mg = cos37° T

 mg = 0.8 90

mg = 72 N

The weight of the object is 72 N

The right answer is option A

Solution 10

As can seen in the figure, distance between the X and Y is 6 unit, distance between the X and Z is 8 unit, and distance between the Y and Z is 10 unit. Let we assume the length of each square is 1 m.

Let we find the force applied to the Y by the X.

 F = k* 5*4 36

 F = 5k 9

Suppose that F2 is the force applied to the charge Z by the charge Y.

 F2 = k* 4*2 100

 F2 = 2k 25

F
=5k
 * 25 2k
9
F2

F
 = 125 18
F2

 F2  = 18F 125

The right answer is option C

The Questions Of This Test

Electrical Force Subject Expression

RISE KNOWLEDGE

January 1 2018

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