Question

A very small object carrying -6.0 μc of charge is attracted to a large, well-anchored, positively charged object. How much kinetic energy does the negatively charged object gain if the potential difference through which it moves is 3.0 mv

Answer 1

Answer:

    Kinetic energy gain by a very small object carrying -6.0 μc of charge is attracted to a large, well-anchored, positively charged object = 18 nJ

Explanation:

  The potential energy gained by negatively charge is the change in potential energy between the points.

 So, Kinetic energy = Change in potential energy = qΔV

 In this case q = 6.0 μc = $6*10^{-6} C$ and ΔV = 3.0 mV = $3*10^{-3} V$

  Substituting

              Kinetic energy gain = $6*10^{-6}*3*10^{-3} = 18*10^{-9}J =18 nJ$

Answer 2

The kinetic energy (KE) gained by the particle is defined as the one-half of the mass of each gas molecule times multiplied by the square of RMS velocity. The kinetic energy gained by the particle will be 18 nJ.

What is kinetic energy?

The kinetic energy (KE) is defined as the one-half of the mass of each gas molecule times multiplied by the square of RMS velocity.

where,

KE is the kinetic energy

m is the mass of each molecule

(V)rms is the RMS velocity

The kinetic energy of the particle is equal to the potential difference between the two charges. The potential energy is given by the product of charge and the potential difference.

$\rm {KE= \frac{1}{2} mv^{2} =qV}$

$\rm {KE= \frac{1}{2} mv^{2} =6\times 10^{-6}\times3\times10^{-3}}$

$\rm {KE= 18\times10^{-9}}\\\\\\KE= 18 nJ$

Therefore the kinetic energy gained by the particle will be 18 nJ.

To learn more about kinetic energy refer to the link ;

brainly.com/question/24134093