Question

.2, Two point charges 30 micrometer and -40 micrometer are held at the origin and at x 0.72 m, respectively particle q = 42 micrometer is released from rest at x 0.28 m. if the initial acceleration of the particle he magnitude of 1 x 10^5 m/s, what is its mass?

Answer 1

As per the question,

$q_1=30\mu C$

$q_2=-40\mu C$

$q_1$ and $q_2$ are at the origin and at $x=0.72m$

$q=42\mu C$

q is at the x= 0.28m

acceleration =$10^5m/sec$

Hence, the net force on the charge q,

$F_{net}=\dfrac{q_1q}{4\pi \epsilon_o r_1^2}+\dfrac{q_2q}{4\pi \epsilon_o r_2^2}$

$F_{net}=\dfrac{9\times 30\times 42\times 10^{-12}\times 10^{9}}{0.28^2}+\dfrac{9\times 40\times 42\times 10^{-12}\times 10^{9}}{0.44^2}$

$F_{net}=144.6+78$

$F_{net}=222.6N$

let the mass is m,

So, $F=ma$

$m=\dfrac{222.6}{10^5}=222.6\times 10^{-5}kg$

or,

$m=2.226gm$