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Question: Two identical particles, each having a charge of \(2.0 \times {10^{ - 4}}{\text{C}}\)and mass of 10g...

Two identical particles, each having a charge of 2.0×10−4C2.0 \times {10^{ - 4}}{\text{C}}and mass of 10g, are kept at a separation of 10 cm and then released. What would be the speeds of the particles when the separation becomes large?

Explanation

Solution

In order to solve this question first we will find the electric force and force due to accelerated motion of particles. Then we will equate them to get the value of acceleration of a particle. Finally we will use Newton's third law of equations to get the required velocity of particles.

Formula used:
Electric force can be expressed as
F=KQr{\mathbf{F}} = \dfrac{{{\mathbf{KQ}}}}{{\mathbf{r}}}
K = Coulomb's constant
Q = charge
R = radius
Using Newton's Second Law of Motion
F = m a
M = mass of an object
A = acceleration of object
Equations of motion
v2=u2+2as{{\mathbf{v}}^2} = {{\mathbf{u}}^2} + {\mathbf{2as}}
V = Final velocity
U = Initial velocity
S = displacement

Complete step-by-step answer:
The acceleration of an object caused by a net force is strictly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object, according to Newton's second law.
Given
Mass = 10g = 10×10−310 \times {10^{ - 3}}kg
r = 10 cm = 10×10−210 \times {10^{ - 2}}m
Electric force can be expressed as
F=KQr{\mathbf{F}} = \dfrac{{{\mathbf{KQ}}}}{{\mathbf{r}}}
Now
K = 9×1099 \times {10^9}
Q = 2.0×10−4C2.0 \times {10^{ - 4}}{\text{C}}
r =10×10−210 \times {10^{ - 2}}m
Substituting all the values, we get
F=KQr=9×109×2×10−410×10−2{\mathbf{F}} = \dfrac{{{\mathbf{KQ}}}}{{\mathbf{r}}} = \dfrac{{{\mathbf{9}} \times {\mathbf{1}}{{\mathbf{0}}^9} \times {\mathbf{2}} \times {\mathbf{1}}{{\mathbf{0}}^{ - 4}}}}{{{\mathbf{10}} \times {\mathbf{1}}{{\mathbf{0}}^{ - 2}}}}
F=1.8×10−7{\mathbf{F}} = 1.8 \times {10^{ - 7}}N
Now using Newton's Second Law of Motion
F = m a
a=1.8×10−710×10−3=1.8×10−3  m/s2{\text{a}} = \dfrac{{1.8 \times {{10}^{ - 7}}}}{{10 \times {{10}^{ - 3}}}} = 1.8 \times {10^{ - 3}}\;{\text{m}}/{{\text{s}}^2}
Using Equations of motion v2=u2+2as{{\mathbf{v}}^2} = {{\mathbf{u}}^2} + {\mathbf{2as}}
v=u2+2as{\mathbf{v}} = \sqrt {{{\mathbf{u}}^2} + {\mathbf{2as}}}
u = initial velocity
a = acceleration
s = displacement
v=0+2+1.8×10−3×10×10−2v = \sqrt {0 + 2 + 1.8 \times {{10}^{ - 3}} \times 10 \times {{10}^{ - 2}}}
v=3.6×10−4=0.6×10−2=6×10−3  m/sv = \sqrt {3.6 \times {{10}^{ - 4}}} = 0.6 \times {10^{ - 2}} = 6 \times {10^{ - 3}}\;{\text{m}}/{\text{s}}
v=6×10−3  m/sv = 6 \times {10^{ - 3}}\;{\text{m}}/{\text{s}}

Note: Make sure you follow the correct unit conversions. There is no other method to solve the problem.
In physics, a force is any interaction that, when unopposed, causes an object to change its motion. A force may cause a mass object to change its velocity (which involves starting to move from a standstill), i.e. accelerate. Intuitively, force can be defined as a push or a pull.