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STALIN [3.7K]
3 years ago
8

A uniform electric field of 2 kNC-1 is in the x-direction. A point charge of 3 μC initially at rest at the origin is released. W

hat is the kinetic energy of this charge at x = 4m?
Physics
1 answer:
ANEK [815]3 years ago
8 0
The electric force acting on the charge is given by the charge multiplied by the electric field intensity:
F=qE
where in our problem q=3 \mu C= 3 \cdot 10^{-6} C and E=2 kN/C=2000 N/C, so the force is
F=(3 \cdot 10^{-6} C)(2000 N/C)=0.006 N

The initial kinetic energy of the particle is zero (because it is at rest), so its final kinetic energy corresponds to the work done by the electric force for a distance of x=4 m:
K(4 m)=W=Fd=(0.006 N)(4 m)=0.024 J
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The Image distance and Magnification of The Image ​will be 30 cm and 3.

<h3>What is focal length?</h3>

The focal length of the lens, which is often expressed in millimeters, is the distance between the lens and the image sensor when the subject is in focus.

Given data;

Focal length,f=?

Image distance,v=?

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Magnification,m= 2.85

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f= 15 Cm

The mirror equation is found as;

\rm \frac{1}{f} =\frac{1}{v} +\frac{1}{u} \\\\\ \frac{1}{15}=\frac{1}{v}+\frac{1}{10} }  \\\\\ v= -30 \ cm

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