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Novay_Z [31]
3 years ago
14

Can someone helppppppppp meeeeeeeeeee plzzzzzz ill give brainliest

Physics
2 answers:
Alecsey [184]3 years ago
4 0
Yes i will help you.
valina [46]3 years ago
3 0

Answer:

<em>exercise 1. </em><em>:</em>

  1. attract ( cuz unlike poles )
  2. repel ( cuz like poles )
  3. repel ( cuz like poles )

<em>exercise</em><em> </em><em>2</em><em>.</em><em> </em><em>:</em>

<em>the</em><em> </em><em>magnetic</em><em> </em><em>pins</em><em> </em><em>are</em><em> </em><em>-</em><em> </em> pin A and pin C

( because only magnets or magnetised materials can repel each other...)

<em> </em><em>i</em><em> </em><em>hope</em><em> </em><em>it</em><em> </em><em>helped</em><em>. </em><em>.</em><em>.</em><em>.</em>

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Two point charges are 10.0cm apart and have charges of 2.0uC and -2.0uC, respectively. What is the magnitude of the electric fie
elena-14-01-66 [18.8K]
The electric field generated by a point charge is given by:
E= k_e \frac{Q}{r^2}
where
k_e = 8.99 \cdot 10^9 Nm^2 C^{-2} is the Coulomb's constant
Q is the charge
r is the distance from the charge

We want to know the net electric field at the midpoint between the two charges, so at a distance of r=5.0 cm=0.05 m from each of them. 

Let's calculate first the electric field generated by the positive charge at that point:
E_1=k_e  \frac{Q_1}{r^2}=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{(2.0 \cdot 10^{-6} C)}{(0.05 m)^2} =+7.19 \cdot 10^6 N/C
where the positive sign means its direction is away from the charge.

while the electric field generated by the negative charge is:
E_2=k_e \frac{Q_1}{r^2}=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{(-2.0 \cdot 10^{-6} C)}{(0.05 m)^2} =-7.19 \cdot 10^6 N/C
where the negative sign means its direction is toward the charge.

If we assume that the positive charge is on the left and the negative charge is on the right, we see that E1 is directed to the right, and E2 is directed to the right as well. This means that the net electric field at the midpoint between the two charges is just the sum of the two fields:
E_{tot} =E_1 + E_2 = 7.19 \cdot 10^6 N/C+7.19 \cdot 10^6 N/C=1.44 \cdot 10^7 N/C
3 0
4 years ago
S When an uncharged conducting sphere of radius a is placed at the origin of an x y z coordinate system that lies in an initiall
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The sphere has a constant potential. It is the electric field.

E = V_{0} = 0

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Outside the sphere, then

V = V_{0} - E_{0}z + \frac{E_{0}a^{3}z}{(x^{2} +y^{2} + z^{2})^{3/2}   }

The elements of the electric field include

E_{x} =\frac{3E_{0}a^{3}xy}{(x^{2} +y^{2} +z^{2})^{5/2}}\\E_{y} = \frac{3E_{0}a^{3}xz}{(x^{2} +y^{2}+z^{2})^{5/2}}

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<h3>In a consistent electric field, is force constant?</h3>

Similar to an ordinary object in the uniform gravitational field near the Earth's surface, a charged item in a uniform electric field experiences a constant force and consequently experiences a uniform acceleration. The vector cross product of p and E determines the torque's direction.

If the charge is positive, the force either moves in the same direction as E or in the opposite direction (if charge is negative).

A torque is experienced by an electric dipole (p) in an even electric field (E). The vector cross product of p and E determines the torque's direction.

To learn more about uniform electric field, visit

brainly.com/question/17426130

#SPJ4

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Answer:

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