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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

01

Figure for net force on charge

Figure is,

02

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" $9.0 脳 10^{9} N 路 m^{2} / C^{2}$ .

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

$F_{1} = 贵肠辞蝉胃 = (\frac{- KqQ}{r^{2}}) 肠辞蝉胃$

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

$F_{2} = - 贵肠辞蝉胃 = - (\frac{KqQ}{r^{2}}) 肠辞蝉胃$

03

Calculation for net force

The term

$肠辞蝉胃 = \frac{a}{r}$

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

$F_{2} = F_{1} = (\frac{- KqQ}{r^{2}}) 肠辞蝉胃$

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

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

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

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

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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Most popular questions from this chapter

What are the strength and direction of the electric field 4.0 cm from a small plastic bead that has been charged to -8.0 nC?

A lightweight metal ball hangs by a thread. When a charged rod is held near, the ball moves toward the rod, touches the rod, then quickly 鈥渇lies away鈥� from the rod. Explain this behavior.

A $2.0 g$ metal cube and a $4.0 g$ metal cube are $6.0 cm$ apart, measured between their centers, on a horizontal surface. For both, the coefficient of static friction is $0.65$ . Both cubes, initially neutral, are charged at a rate of $7.0 nC / s$ . How long after charging begins does one cube begin to slide away? Which cube moves first?

What is the force $\overset{鈫�}{F}$ on the $1.0 nC$ charge at the bottom in FIGURE $P 22.46$ $?$ Give your answer in component form.

Two 3.0 g point charges on 1.0-m-long threads repel each other after being equally charged, as shown in FIGURE CP22.73. What is the magnitude of the charge q?

See all solutions

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