An object with a charge of -4.0 μc and a mass of 1.4×10−2 kg experiences an upward electric force, due to a uniform electric field, equal in magnitude to its weight. part a find the magnitude of the electric field. express your answer using two significant figures.

Respuesta :

The formula that we will be using is:

Felec = qE = mg => E = mg/q

So using the formula of E:
E = (0.014kg)(9.8 m/s^2) / -4.0*10^-6C = 34,000 N/C

This is the magnitude of the field. The direction must be downward to give an upward force on a negative charge.

Answer:

34000N/C (to 2sf)

Explanation:

Using the relationship F = qE where F is the electric force experienced by the object

q is the charge on the object

E is the electric field.

If the body experiences an upward electric force, due to a uniform electric field, equal in magnitude to its weight W then F = W

The equation above can be rewritten as W = qE ... 2

From equation 2;

E = W/q

Since W = mass m × acceleration due to gravity g

E = mg/q

Given the data

Mass = 1.4×10^−2 kg

g = -9.81m/s²(since it experiences an upward force)

q = -4.0 μc = -4×10^-6C

E = (1.4×10^−2 × -9.81)/-4×10^-6

E = 3.43×10^(-2-(-6))

E = 3.43×10^(-2+6)

E = 3.43 × 10^4

E = 34300

E = 34000 N/C (to 2sf)

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