Question

Three point charges lie along a straight line as shown in the figure below, where q1...

uploaded imageThree point charges lie along a straight line as shown in the figure below, where q1 = 5.94 µC, q2 = 1.41 µC, and q3 = -2.2 µC. The separation distances are d1 = 3.00 cm and d2 = 2.00 cm. Calculate the magnitude and direction of the net electric force on each of the charges.

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Answer #1
Concepts and reason

The concepts used to solve this problem is electric field magnitude and direction.

Initially, the principle to be applied to this question is the “Principle of Superposition”. Applying this principle on each of the charges, to find out the magnitude and direction of the force on the charge due to the other two charges.

Coulomb’s Law” is to be used to find the magnitude of the electric force between the charges. From the expression of the Coulomb force, it depends on the charge and the distance between the charges.

Fundamentals

The principle of superposition states that the force between two charges is totally independent of the influence of the other charges. If the force on a charge say due to being and due to be and so on, then the resultant force on is

Here, N
is the total number of charges.

The formula for electric force is as follows:

kq,q2
2

Here, k
is constant and the value ofk
is equal tok 8.99*10°N-m2 C
, is the magnitude of charge and is the distance between the two charges.

Calculate the magnitude of charge in between the charge and42
.

The expression of the magnitude of charge in between and42
is equal,

12
2

Here,12
is the force between the charge and42
.

Substitute8.99X10°N m2-C2
fork
,5.94 C
for,1.41 μC
for42
and 3.00 cm
ford
in the above expression of the force.

(8.99 x 10N-m2-C2(5.94 C)(1.41 /C)
(3.00 cm)
(8.99x10N-m2-C2)(5.94 /iC) 1x10° C
12
C
C
1x10 C
2
m
(3.00 cm
100 cm
=83.66 N

From the above expression of the force, both charges are the same polarity, therefore, the nature of the force is repulsive.

Calculate the magnitude of charge in between the charge and42
.

The expression of the magnitude of charge in between and42
is equal,

(d,+d,
2

Substitute8.99X10°N m2-C2
fork
,5.94 C
for,2.2 μC
foran(3.00 cm+2.00 cm)
ford+d,
in the above expression of the force.

(8.99х10°N-m? -C?)(5.94 дС)(2.2 дС)
(3.00 ст +2.00 спm)*
13
дезию(,
(8.99х10° N-m*-C)(5.94 AC)x10° иС
C
1x10° дС
1m
(5.00 сm

From the above expression of the force, one charge has positive polarity and another charger has negative polarity. Therefore, the nature of the force is attractive.

The expression of the net force on charge is equal to,

12
13

Substitute83.66 N
for12
and|46.99 N
for13
in the above expression of the net forceF83.66 N-46.99 N
= 36.67 N

The direction of the force is repulsive.

Calculate the magnitude of charge in between the charge 42
and.

The expression of the magnitude of charge in between and is equal,

kq2qs
23
2
(d,)

Substitutefor,1.41 μC
for42
,foran2.00 cm
forin the above expression of the force.

(8.99x10N m2 C2)(1.41 uC)(2.2 /C)
F23
(2.00 cm)
1 C
(8.99*10°N-m2 -C* )(1.41 //C)\1x10* C
1 C
1x10 C
1m
(2.00 cm
100 cm
=69.

The resultant force on 42
is towards the right.

The expression of the net force on charge 42
is equal to,

12
23

Substitute forand69.71 N
forF,
23
in the above expression of the net force,

=83.66 N 69.71 N
= 153.37 N

The expression of the net force on the charge is equal to,

13
23

Substituteforandfor in the above expression of the net force

46.99 N 69.71 N
=116.7 N

The resultant force on is towards the left.

Ans:

The magnitude of the force on the charge it is 36.67 N
and the direction is west.

The magnitude of the force on the charge 42
it is 153.37 N
and the direction is west.

The magnitude of the force on is equal to 115.52N
and the direction is west.

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