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kotykmax [81]
3 years ago
13

A positively charged particle 1 is at the origin of a Cartesian coordinate system, and there are no other charged objects nearby

. You need the electric field magnitude at the position (1.00 nm , 8.00 nm ) to be zero, so that no electric force is exerted on any charged object placed at that location.
Where would you need to place particle 2 if it carried the same charge as particle 1?

Physics
1 answer:
8090 [49]3 years ago
3 0

Answer:

P=(2 nm, 8mn)

Explanation:

Given :

Position of positively charged particle at origin, O=(0\ nm,0\ nm)

Position of desired magnetic field, D\equiv(1\ nm,8\ nm)

Magnitude of desired magnetic field, E=0\ N.C^{-1}

Let q be the positive charge magnitude placed at origin.

<u>We know the distance between the two Cartesian points is given as:</u>

d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}

<u>For the electric field effect to be zero at point D we need equal and opposite field at the point.</u>

\frac{1}{4\pi.\epsilon_0} \times \frac{q}{r^2 } =\frac{1}{4\pi.\epsilon_0} \times \frac{q}{r^2 }

\therefore (1-0)^2+(8-0)^2=r^2

r^2=65\ nm

r=\sqrt{65}

as we know that the electric field lines emerge radially outward of a positive charge so the second charge will be at equally opposite side of the  given point.

assuming that the second charge is placed at (x,y) nano-meters.

Therefore,

x=2\times 1=2\ nm

and

y=2\times 8=16\ nm

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Answer:

The same as the escape velocity of asteorid A (50m/s)

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