Question

A charge particle having charge 4 × 10−3 C moves from state of rest in uniform electric field of magnitude 5 × 103 V/m. What is its KE after 10 second? (m = 2 × 10−9 kg)​

Answer 1

Given:

A charge particle having charge 4 × 10^(−3) C moves from state of rest in uniform electric field of magnitude 5 × 10³ V/m.

To find:

KE after 10 seconds?

Calculation:

Acceleration of the charge be 'a':

$\rm \: a = \dfrac{force}{mass}$

$\rm \implies a = \dfrac{Eq}{m}$

$\rm \implies a = \dfrac{5 \times {10}^{3} \times 4 \times {10}^{ - 3} }{2 \times {10}^{ - 9} }$

$\rm \implies a = {10}^{10} \: m {s}^{ - 2}$

Now, velocity after 10 seconds:

$\rm \: v = u + at$

$\rm \implies \: v = 0 + ( {10}^{10} \times 10)$

$\rm \implies \: v = {10}^{11} \: m {s}^{ - 1}$

Now, Kinetic Energy is:

$\rm KE = \dfrac{1}{2} m {v}^{2}$

$\rm \implies KE = \dfrac{1}{2} \times (2 \times {10}^{ - 9} ) \times {10}^{11}$

$\rm \implies KE = 100 \: joule$

So, kinetic energy is 100 Joule.