Question

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?

Answer 1

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.

We know the distance between the two Cartesian points is given as:

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

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

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

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

$r^2=65\ nm$

$r=\sqrt{65}$

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

$y=2\times 8=16\ nm$

Answer 2

The position of second charge is (2,16).

• Given: In the Cartesian coordinate system the positive charge is on the origin and the point p is at the (1,8) point, that is the X and Y coordinate plane.
• The electric force (F1) of charge one at a point p is repulsive.
• Suppose the distance between the charge one and the point p is d.
• To make the electric force zero at point p a positive charge is required to be placed at a distance d and in the direction of F1.
• Thus, the positive of the second charge will be at (2,16) on the Cartesian coordinate plane.

Learn more about coordinate system:

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