Question

What is the angle? (two point charges on 1.0m long threads repel each other after being charged...

What is the angle?

Two 5.0g point charges on 1.0m long threads repel each other after being charged to +100 nC, as shown in the figure.

What is the angle theta?You can assume that theta is a small angle.

The answer is 4.4 degree.Please give steps to getting the answer.

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Answer #1
Concepts and reason

The concepts required to solve the problem are the force equilibrium, trigonometric relations and electrostatic force.

Initially, consider the components of tension on the thread. Later, use the trigonometric relations to simplify the equation for tension. Finally, compare the force component with equivalent forces to calculate the angle.

Fundamentals

The tension on a thread which held a mass at an angle from the vertical axis has two components. The horizontal component of the tension is,

T = T sino

Here, is the tension, is the angle from the vertical axis.

The vertical component of the tension is,

T, = T cos

The weight of an object is,

W = mg

Here, is the mass of the object and is the acceleration due to gravity.

The electrostatic force between two identical charges is,

Here, is the Coulomb constant, is the charge of each charged particle and is the distance between the charges.

For small angle , sin
= 0
and cos O=1
.

The opposite side of an angle in a right-angle triangle is,

d=Isin

Here, is the length of hypotenuse.

The tension on the thread hanging two charged particles have two components of force on it.

The horizontal component of the tension is,

T = T sin

The horizontal component of the force acting on the particle is equivalent to the electrostatic force between them. The electrostatic force between the charges is,

The distance between the charges is,

r=21 sine

Substitute 21 sin
for in the expression for electrostatic force . The electrostatic force between the point charges is,

F=k
(21 sin o)

The horizontal component of tension on the thread is,

T = F

Substitute T sin
for and k9²
(21 sin o)
for . The horizontal component of force is,I sin (=k
(21 sin o)

The angle the charges makes with the vertical axis is very small. Therefore,

sin
= 0

Substitute or . The horizontal component of tension on the thread is,

TO=k_9
4102

The vertical component of tension is,

T = T cos

The vertical component of the tension on the thread is balanced by the weight of the charged particle. The weight of charged particle is,

W = mg

The vertical component of tension is,

T=W

Substitute T cose
for and for . The vertical component of force is,

T cos 0 = mg

The angle the charges makes with the vertical axis is very small. Therefore,

cos O=1

Substitute for . The vertical component of the tension on the thread is,

T = mg

The vertical component of tension on the thread due to the charges is,

T = mg

The horizontal component of tension on the thread due to the charges is,

Rewrite the equation for horizontal component of the force to calculate the angle. The angle is,

Consider the vertical component of the tension. Substitute for to find the angle. The angle is,

0= k 41 mg

Substitute 9.0x10° N-m®/C
for , 100 nC
for , for , for ,and 9.8m/s2
forand solve for .

(9.0x10° N·m²/C) —
(100me) ( 10 )
10-kg) 98 m/s2)
= 7.7x10-2 rad
=7.7x10-2 radx
1 rad
=4.4°

Therefore, the angle is .

Ans:

The angle is equal to .

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