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slavikrds [6]
4 years ago
12

Suppose a negative point charge is placed at x = 0 and an electron is placed at some point P on the positive x-axis. What is the

direction of the electric field at point P due to the point charge, and what is the direction of the force experienced by the electron due to that field?
O E along -X; F along –X
O E along +x; F along +x
O E along –x; F along +x
O E along +x; F along - x
Physics
1 answer:
jolli1 [7]4 years ago
3 0

Answer:

E along –x; F along +x

Explanation:

  • When a negative point charge is placed at x=0 and an electron is place at any point P on the positive x-axis the as we know that the like charges repel each other, but there will be no change in the natural tendency of the individual electric field lines.
  • So the direction of the electric field lines at point P due to the point charge will be towards the negative x-axis.
  • The direction of force on the electron due to the electric field of point charge at x=0 will be towards positive x-axis in accordance of the repulsion effect.
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My definition: The meaning or deintion of any word.

5 0
4 years ago
In which city rusting is a problem delhi or mumbai
ss7ja [257]

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3 0
3 years ago
As a train accelerates uniformly, it passes successive 5-kilometer markers while traveling at velocities 10 m/s and 25 m/s. what
gregori [183]
<span>122.0 km/hr. First let’s make sure all of our units are in the base meter form: i.e. convert 5km to 5000m. (We will convert back to km later). The first thing to do is look at the equation relating velocity, acceleration, and distance: Vf^2 = Vi^2 + 2*a*d, where Vf is final velocity, Vi is initial velocity, a is acceleration, and d is distance. 25^2 = 10^2 + 2*a*5000 =?> 625 = 100 +10000a => a= 0.0525m/s^2. Now that we have acceleration, we can use the same equation again with different numbers.: Vf^2 = Vi^2 + 2*a*d = 25^2 + 2*0. 0525m*5000 = 625 + 525 =1150 => Vf^2 = 1150 => 33.9m/s. Convert to km/hour: 33.9m/s * 1km/1000m *60s/1min * 60min/ 1 hr = 122.0 km/hr.</span>
8 0
4 years ago
A speck of dust with mass 12 mg and electric charge 10 μC is released from rest in a uniform electric field of magnitude 850 N/C
Marta_Voda [28]

Answer:

a=708.3m/s^2

Explanation:

The force experimented by a charge <em>q </em>in a uniform electric field <em>E</em><em> </em>is <em>F=qE</em>.

Newton's 2nd Law tells us that the relation between acceleration <em>a</em> a mass <em>m </em>experiments when a force <em>F </em>is applied to it is <em>F=ma</em>.

Combining these equations we have <em>am=qE</em>, and since we want the acceleration of the speck of dust, we substitute our values:

a=\frac{qE}{m}=\frac{(10\times10^{-6}C)(850N/C)}{12\times10^{-6}Kg}=708.3m/s^2

4 0
3 years ago
Mixing baking soda with vinegar is an example of an oxidation-reduction
34kurt

Answer:

B. False

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