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3 years ago

What is the electric force acting between two charges of -0.0050 C and

Physics

1 answer:

love history [14]3 years ago

6 0

The electric force is $-3.6\cdot 10^8 N$ (attractive)

Explanation:

The magnitude of the electric force between two charges is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=9\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

r is the separation between the two charges

In this problem, we have the following:

$q_1 = -0.0050 C$ (charge 1)

$q_2 = +0.0050 C$ (charge 2)

r = 0.025 m (distance)

Substituting, we find the electric force between the two charges:

$F=(9\cdot 10^9) \frac{(-0.0050)(0.0050)}{(0.025)^2}=-3.6\cdot 10^8 N$

And the negative sign means the force is attractive, since the two charges have opposite sign.

Learn more about electric force:

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Answer:

The required angle is (90-25)° = 65°

Explanation:

The given motion is an example of projectile motion.

Let 'v' be the initial velocity and '∅' be the angle of projection.

Let 't' be the time taken for complete motion.

Let 'g' be the acceleration due to gravity

Taking components of velocity in horizontal(x) and vertical(y) direction.

$v_{x}$ = v cos(∅)

$v_{y}$ = v sin(∅)

We know that for a projectile motion,

t = $\frac{2vsin(∅)}{g}$

Since there is no force acting on the golf ball in horizonal direction.

Total distance(d) covered in horizontal direction is -

d = $v_{x}$ ×t = vcos(∅)× $\frac{2vsin(∅)}{g}$ = $\frac{v^{2}sin(2∅) }{g}$ .

If the golf ball has to travel the same distance 'd' for same initital velocity v = 23m/s , then the above equation should have 2 solutions of initial angle 'α' and 'β' such that -

α +β = 90° as-

d = $\frac{v^{2}sin(2α) }{g}$ = $\frac{v^{2}sin(2[90-β]) }{g}$ = $\frac{v^{2}sin(180-2β) }{g}$ = $\frac{v^{2}sin(2β) }{g}$ .

∴ For the initial angles 'α' or 'β' , total horizontal distance 'd' travelled remains the same.

∴ If α = 25° , then

β = 90-25 = 65°

∴ The required angle is 65°.

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