Question

A speck of dust with mass 12 mg and electric charge 10 μC is released from rest in a uniform electric field of magnitude 850 N/C. Calculate the acceleration of the dust speck, ignoring gravitation. ______ m/s^2

Answer 1

Answer:

$a=708.3m/s^2$

Explanation:

The force experimented by a charge q in a uniform electric field E is F=qE.

Newton's 2nd Law tells us that the relation between acceleration a a mass m experiments when a force F is applied to it is F=ma.

Combining these equations we have am=qE, and since we want the acceleration of the speck of dust, we substitute our values:

$a=\frac{qE}{m}=\frac{(10\times10^{-6}C)(850N/C)}{12\times10^{-6}Kg}=708.3m/s^2$