Question

What must the charge (sign and magnitude) of a 1.70 g particle be for it to remain balanced against gravity when placed in a downward-directed electric field of magnitude 700 N/C ?

Answer 1

Answer:

Charge, $q=-2.38\times 10^{-5}\ C$

Explanation:

Mass of the particle, m = 1.7 g = 0.0017 kg

Electric field, E = 700 N/C

We need to find the charge particle to  remain balanced against gravity when placed in a downward-directed electric field. So,

q E = m g

$q=\dfrac{mg}{E}$

$q=\dfrac{0.0017\ kg\times 9.8\ m/s^2}{700\ N/C}$

q = 0.0000238 C

$q=-2.38\times 10^{-5}\ C$

So, the magnitude of charged particle is $2.38\times 10^{-5}\ C$. As the direction of electric field is in downward direction, so the charge is negative. Hence, this is the required solution.