Question

Consider a charged particle of mass 0.004~kg0.004 kg placed in a downward-pointing electric field of magnitude 700~\text{N/C}700 N/C. Determine both the sign and magnitude of the particle's net charge that allows it to "levitate" motionlessly against gravity.

Answer 1

Answer:

The sign of the charge is negative and the magnitude of the charge is $5.6\times 10^{- 5}\ C$

Solution:

Mass of the particle, m = 0.004 kg

Electric Field, E = 700 N/C

Now,

• The force acting on the charged particle in the downward acting electric field must be in the opposite direction to that of the electric field.

• Since the direction of the field is always from positive to negative and the direction of the force is opposite to that of the field, the nature of the charge is negative.

To calculate the magnitude of the charge:

Force due to the mass of the charged particle:

F = mg

The force due to electric field:

$F_{E} = qE$

where

q = charge

Thus to maintain the equilibrium:

$F = F_{E}$

$mg = qE$

$|q| = \frac{mg}{E}$

the magnitude of the charge:

$|q| = \frac{0.004\times 9.8g}{700} = 5.6\times 10^{- 5}\ C$

Since, the nature of the charge is negative:

$q = - 5.6\times 10^{- 5}\ C$