Question

1)What is the force on the 1.0nC charge in the figure?(Figure 1) Give the magnitude. 2)Give...

1)What is the force on the 1.0nC charge in the figure?(Figure 1) Give the magnitude.

2)Give the direction.

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Answer #1
Concepts and reason

The required concepts to solve the problem are Coulomb force and vector algebra.

First, using the Coulomb force, find the net force acting on1.0nC
due to the two2.OnC
charges. Then, using vector algebra, find the direction of the net force acting on1.0nC
.

Fundamentals

The force between two charges separated by a distance is called the Coulomb force. The Coulomb force is,

bby

Here, is the Coulomb constant, is the first charge, is the second charge andis the distance between the charges.

A vector is a physical quantity with both magnitude and direction. A vector can be split into a vertical component and a horizontal component. If the angle of the vector is measured with the horizontal axis, then thecomponent is the horizontal component and thecomponent is the vertical component, whereis the angle made by the vector with the horizontal axis.

The force on charge1.0nC
due to the charge2.OnC
on the left is,

kq,92
F =

Substitute9x10° Nm²/C
for, 1.0nC
for, 2.OnC
forand1.0cm
for.

(9x10 Nm? (C)(1.0nc) ( 90 )(2010)( me
= 1.8x10N

The force on charge1.0nC
due to the charge2.OnC
on the right is,

kq,92
F,=?

Substitute9x10° Nm²/C
for, 1.0nC
for, 2.OnC
forand1.0cm
for.

InC)
F =
(9*10*Nm(C)(.ne) ( 1 12.0nc) ( vom
(1.0cm) by my
10-2m
Icm
= 1.8x10N

Both the forces are vectors. As both the charges are positive, the forceacts along the line joining charge1.0nC
and the charge 2.OnC
on the left away from the charge. The forceacts along the line joining charge1.0nC
and the charge 2.OnC
on the light away from the charge.

By splitting these vectors into components, the horizontal components are equal and opposite, and they get cancelled. The vertical components get added.

Since, the net force acting on1.0nC
is,

F = FsinO+F, sino

Substitute 1.8x104N
forandandfor.

F = (1.8x104N)sin 60°+(1.8x10* N)sin 60°
= 2(1.8x10+N)sin 60°
= 3.12x104N

As both the charges are positive, the forceacts along the line joining chargeand the charge on the left away from the charge. The forceacts along the line joining chargeand the charge on the light away from the charge.

By splitting the vectors into components, thecomponent ofwill be in the positivedirection or towards right andcomponent ofwill be in the positivedirection or upwards.

= J(F cose) +(F, sin o)

Thecomponent ofwill be in the negativedirection or towards right andcomponent ofwill be in the positivedirection or upwards.

F} = (F, coso)? +(F, sin o)

As thecomponents are equal and opposite, they get cancelled. The remainingcomponents are equal, and both directed vertically upwards.

Therefore, the direction of the net force acting onis vertically upwards.

Ans:

The net force acting on1.0nC
is 3.12x10 N
.

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