Question

Find n

$r = \frac{2\sqrt{5} \times 10^{-3}}{3} m$ $Q = \frac{\sqrt{20}}{9} MC$ $mass = 3Kg$

Find initial acceleration of both initial acceleration of +Q is $\frac{n}{3} \times 10^{-9} m/s^2$

• A. 10^18
• B. 10^9
• C. 3*10^18
• D. 3*10^9

Answer 1

Answer: (A)

10^18

Answer 2

The electrostatic force between two charges $+Q$ and $-Q$ separated by distance $r$ is $F = kQ^2/r^2$. The acceleration of a charge is $a = F/m$. By substituting the given values of $Q$, $r$, and $mass$, and assuming $Q$ is in milli Coulombs, the force is calculated as $10^9 N$. The acceleration of the charge $+Q$ is then $10^9/3 m/s^2$. Comparing this with the given format $\frac{n}{3} \times 10^{-9} m/s^2$, we find $n = 10^{18}$.