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3 years ago

Which statement is true of the electric field at a distant from the source of charge?

Physics

1 answer:

iVinArrow [24]3 years ago

8 0

Answer:

A negative charge, if free to move in an electric field, will move from a low potential point to a high potential point. To move a positive charge against the electric field, work has to be done by you or a force external to the field.

Explanation:

Mark as Brainliest plz!!!

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Margaret is the new director of research at a well-known pharmaceutical company. She has been asked to design a set of research

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Answer:

Double blind experiment

Explanation:

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Hence , the new director Margaret , need to design a set of double - blind experiments.

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3 years ago

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3 years ago

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timurjin [86]

Answer:

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3 years ago

Three identical charges q form an equilateral triangle of side a with two charges on the x-axis and one on the positive y-axis.

shusha [124]

Answer:

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

Explanation:

Given:

- Three identical charges q.

- Two charges on x - axis separated by distance a about origin

- One on y-axis

- All three charges are vertices

Find:

- Find an expression for the electric field at points on the y-axis above the uppermost charge.

- Show that the working reduces to point charge when y >> a.

Solution

- Take a variable distance y above the top most charge.

- Then compute the distance from charges on the axis to the variable distance y:

$r = \sqrt{(\frac{\sqrt{3}*a }{2} + y)^2 + (a/2)^2 }$

- Then compute the angle that Force makes with the y axis:

cos(Q) = sqrt(3)*a / 2*r

- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:

F_1,2 = 2*F_x*cos(Q)

- The total net force would be:

F_net = F_1,2 + kq / y^2

- Hence,

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

- Now for the limit y >>a:

$F_n = k*q*(\frac{2*y(1 + \frac{\sqrt{3}*a }{2*y}) }{y^3((1+ \frac{\sqrt{3}*a }{2*y})^2 + (a/y*2)^2)^1.5 }) +\frac{1}{y^2} )$

- Insert limit i.e a/y = 0

$F_n = k*q*(\frac{2}{y^2} +\frac{1}{y^2}) \\\\F_n = 3*k*q/y^2$

Hence the Electric Field is off a point charge of magnitude 3q.

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3 years ago

A baseball pitcher throws a ball at 40 ms^-1. if the acceleration is approximately constant over a distance of 2 m, how large is

Reil [10]

We have the equation of motion $v^2=u^2+2as$ , where v i the final velocity, u is the initial velocity, a is the acceleration and s is the displacement

Here final velocity, v = 40m/s

Initial velocity, u = 0 m/s

Displacement s = 2 m

Substituting $40^2=0^2+2*a*2\\ \\ a=400m/s^2$

So the baseball pitcher accelerates at 400m/ $s^2$ to release a ball at 40 m/s.

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3 years ago