A charged particle of mass 6.38x10^-31kg is suspended motionless in the air by an electric field. If the electric field is 7.52x10^-21 N/C downwards, what is the magnitude and sign of the charge of

the particle?

Require than m * g = q * E

q is the charge. m is the mass in kg.
g is the acceleration of gravity, in m/s^2.
E is the field strength, in N/C
Solve for the field, charge, q, in Coulombs.

The charge must be negative so that the upward electrical force can balance the weight.

To determine the magnitude and sign of the charge of the particle, we can use the formula for the force experienced by a charged particle in an electric field.

The formula is given by:
F = qE

Where:
F is the force experienced by the particle,
q is the charge of the particle,
E is the electric field strength.

In this case, the force F is equal to zero because the particle is suspended motionless in the air. So we have:

0 = qE

To find the charge q, we rearrange the equation to solve for q:

q = 0 / E = 0

This means that the charge of the particle is zero. Therefore, the magnitude of the charge is zero, and the particle is electrically neutral.