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3 years ago

Is it possible for the electric field between two positive charges to equal zero along the line joining the two charges?

Physics

1 answer:

MrRa [10]3 years ago

6 0

Answer:

Yes electric field between the line joining may be zero when both the charges are of opposite magnitude

Explanation:

Electric field due to a point charge is given by $E=\frac{1}{4\pi \epsilon _0}\frac{Q}{r^2}$

From the relation of magnitude of electric field we can see that magnitude of electric field depends upon the m,magnitude of charge and distance between them

So electric field between the line joining may be zero when both charges are of opposite magnitude ( Means if one charge is positive then other is negative )

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3 years ago

You are traveling at 100 km/hr to make it to work on time. You have to be to work in .25 hr or you will be late. You have 16 km

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Answer: Yes

Explanation:

Velocity $V$ is defined as the distance traveled $d$ in a specific time $t$ :

$V=\frac{d}{t}$

If you are traveling at $V=100 km/h$ a distance $d=16 m$ , then the time it will take you to be at work is:

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3 years ago

A student pushes a 0.2 kg box against a spring causing the spring to compress 0.15 m. When the spring is released, it will launc

german

Answer:

The maximum height the box will reach is 1.72 m

Explanation:

F = k·x

Where

F = Force of the spring

k = The spring constant = 300 N/m

x = Spring compression or stretch = 0.15 m

Therefore the force, F of the spring = 300 N/m×0.15 m = 45 N

Mass of box = 0.2 kg

Work, W, done by the spring = $\frac{1}{2} kx^2$ and the kinetic energy gained by the box is given by KE = $\frac{1}{2} mv^2$

Since work done by the spring = kinetic energy gained by the box we have

$\frac{1}{2} mv^2$ = $\frac{1}{2} kx^2$ therefore we have v = $\sqrt{\frac{kx^2}{m} }$ = $x\sqrt{\frac{k}{m} }$ = $0.15\sqrt{\frac{300}{0.2} }$ = 5.81 m/s

Therefore the maximum height is given by

v² = 2·g·h or h = $\frac{v^2}{2g}$ = $\frac{5.81^{2} }{2*9.81}$ = 1.72 m

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3 years ago