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lianna [129]
3 years ago
14

You are given two unknown point charges, Q1, and Q2. At a point on the line joining them, one third of the way from Q1 to Q2 the

electric field is zero. What can you say about these two charges?
Physics
2 answers:
Tems11 [23]3 years ago
8 0

Answer:

  • The charges are of the same sign
  • The charge Q2 its four times the charge Q1

Explanation:

  • The charges are of the same sign.

If they were of opposite signs, in the line joining them, the electric field will not be zero at any point, as in all points of this line will be pointing towards the negative charge.

  • The charge Q2 its four times the charge Q1.

The equation for the electric field produced by a point charge its:

E = \frac{1}{4 \pi \epsilon_0} \frac{Q}{d^2}

taking in consideration that the electric field its a vector that points radially outward from the point charge, and the equation above only gives us its length.

Now, If we got a zero electric field, that must mean that the electric fields from both charges are, in magnitude, equals. This means:

\frac{1}{4 \pi \epsilon_0} \frac{Q_1}{d_1^2}=\frac{1}{4 \pi \epsilon_0} \frac{Q_2}{d_2^2}

\frac{Q_1}{d_1^2} = \frac{Q_2}{d_2^2}

We also now that the distance to Q1 from this point its one third the distance from Q1 to Q2, this must mean that d_2 its 2 times d_1.

\frac{Q_1}{d_1^2} = \frac{Q_2}{ (2 * d_1)^2}

We can get ride of the d_1 factor that appears on both sides:

\frac{Q_1}{d_1^2} = \frac{Q_2}{ 2^2 d_1^2}

\frac{1}{d_1^2} Q_1 = \frac{1}{d_1^2}\frac{Q_2}{ 4 }

Q_1 = \frac{Q_2}{ 4 }

And we finally get:

Q_2 = 4 * Q_1

podryga [215]3 years ago
3 0
The formula that is applicable here is E = kQ/r^2 in which the energy of attraction is proportional to the charges and inversely proportional to the square of the distance. In this case, 
kQ1/(r1)^2 = kQ2/(r2)^2  r1=l/3, r2=2l/3solve Q1/Q2
kQ1/(l/3)^2 = kQ2/(2l/3)^2  kQ1/(l^2/9) = kQ2/(4l^2/9)Q1/Q2 =   1/4
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Answer:

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Given:

mass of the dog = 14.3 kg

To find:

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Formula used:

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W = mg

Where, W = weight of the dog

m = mass of the dog

g = acceleration due to gravity

Solution:

Weight of the dog is given by,

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W = 14.3 × 9.8

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Weight of the dog on surface of earth is 140.14 Newton.


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A. How far does a 100-newton force have to move to do 1,000 joules
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Work done by a force is given as the product of force and the distance moved by the force.

<h3>What is work done?</h3>

Work done is the product of force and the distance moved by the the force.

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Thus, distance required by the 100 N force is given as:

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A projectile is launched straight up from a height of 960 feet with an initial velocity of 64 ft/sec. Its height at time t is h(
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Answer:

a) t=2s

b) h_{max}=1024ft

c) v_{y}=-256ft/s

Explanation:

From the exercise we know the initial velocity of the projectile and its initial height

v_{y}=64ft/s\\h_{o}=960ft\\g=-32ft/s^2

To find what time does it take to reach maximum height we need to find how high will it go

b) We can calculate its initial height using the following formula

Knowing that its velocity is zero at its maximum height

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

0=(64ft/s)^2-2(32ft/s^2)(y-960ft)

y=\frac{-(64ft/s)^2-2(32ft/s^2)(960ft)}{-2(32ft/s^2)}=1024ft

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a) From the equation of height we calculate how long does it take to reach maximum point

h=-16t^2+64t+960

1024=-16t^2+64t+960

0=-16t^2+64t-64

Solving the quadratic equation

t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}

a=-16\\b=64\\c=-64

t=2s

So, the projectile reach maximum point at t=2s

c) We can calculate the final velocity by using the following formula:

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

v_{y}=±\sqrt{(64ft/s)^{2}-2(32ft/s^2)(-960ft)}=±256ft/s

Since the projectile is going down the velocity at the instant it reaches the ground is:

v=-256ft/s

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