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

Charge,
`q=-2.38* 10^(-5)\ C`

Step-by-step 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=(mg)/(E)`

`q=(0.0017\ kg* 9.8\ m/s^2)/(700\ N/C)`

q = 0.0000238 C

`q=-2.38* 10^(-5)\ C`

So, the magnitude of charged particle is
`2.38* 10^(-5)\ C`. As the direction of electric field is in downward direction, so the charge is negative. Hence, this is the required solution.