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krek1111 [17]
2 years ago
13

Electric potential difference is the change in kinetic energy per unit charge in an electric field.

Physics
1 answer:
777dan777 [17]2 years ago
3 0

Hi there!

<u>False. </u>

Electric potential difference is the change in ELECTRIC POTENTIAL ENERGY per unit charge.

\large\boxed{\Delta V = \frac{\Delta U}{q}}

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