Chapter 22: Q. 50 (page 626)

A positive point charges $Q$ located at $x = a$ and a negative point charge $- Q$ is at $x = - a$ .A positive charge $q$ can be place anywhere on the $y$ - axis. Find an expression for ${(F_{net})}_{x}$ , the component of the net force on $q$ .

Short Answer

Expert verified

The component of the net force on charge is $(\frac{- 2 KqQa}{{(y^{2} + a^{2})}^{(3 / 2)}})$ .

Step by step solution

Figure for net force on charge

Figure is,

Calculation for force

The two particles exert forces on each other, with the force decreasing as the distance between them grows and increasing as the size of the charges increases.

$F_{C} = F_{1 on 2} = F_{2 on 1} = \frac{K |q_{1}| |q_{2}|}{r^{2}}$

The electrostatic constant is localid="1649101048148" $9.0 \times 10^{9} N \cdot m^{2} / C^{2}$ .

The charge $q$ is a long way away.

$r = \sqrt{y^{2} + a^{2}}$

$- Q$ and $+ Q$ from each charge It's on the $y$ - axis as well.

As a result, the force between $q$ and $Q$ is equal to

$F_{1} = Fcosθ = (\frac{- KqQ}{r^{2}}) cosθ$

The force between $q$ and $Q$ can be calculated as follows:

$F_{2} = - Fcosθ = - (\frac{KqQ}{r^{2}}) cosθ$

Calculation for net force

The term

$cosθ = \frac{a}{r}$

Two forces $F_{1}$ and $F_{2}$ :

$F_{2} = F_{1} = (\frac{- KqQ}{r^{2}}) cosθ$

$= (\frac{- KqQ}{r^{2}}) (\frac{a}{r})$

$= (\frac{- KqQa}{r^{3}})$

$= (\frac{- KqQa}{{(\sqrt{y^{2} + a^{2}})}^{3}})$

$= (\frac{- KqQa}{{(y^{2} + a^{2})}^{(3 / 2)}})$

Net force is,

$F_{net} = F_{1} + F_{2}$

$= (\frac{- KqQa}{{(y^{2} + a^{2})}^{(3 / 2)}}) + (\frac{- KqQa}{{(y^{2} + a^{2})}^{(3 / 2)}})$

$= (\frac{- 2 KqQa}{{(y^{2} + a^{2})}^{(3 / 2)}})$

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A positive point charges $Q$ located at $x = a$ and a negative point charge $- Q$ is at $x = - a$ .A positive charge $q$ can be place anywhere on the $y$ - axis. Find an expression for ${(F_{net})}_{x}$ , the component of the net force on $q$ .

Short Answer

Step by step solution

Figure for net force on charge

Calculation for force

Calculation for net force

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A positive point chargesQlocated atx=aand a negative point charge -Qis atx=-a.A positive chargeqcan be place anywhere on they- axis. Find an expression forFnetx, the component of the net force onq.

Short Answer

Step by step solution

Figure for net force on charge

Calculation for force

Calculation for net force

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Magnetism and Electromagnetic Induction

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Study anywhere. Anytime. Across all devices.

A positive point charges $Q$ located at $x = a$ and a negative point charge $- Q$ is at $x = - a$ .A positive charge $q$ can be place anywhere on the $y$ - axis. Find an expression for ${(F_{net})}_{x}$ , the component of the net force on $q$ .