Respuesta :
Answer:
The sign of the charge is negative and the magnitude of the charge is [tex]5.6\times 10^{- 5}\ C[/tex]
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:
[tex]F_{E} = qE[/tex]
where
q = charge
Thus to maintain the equilibrium:
[tex]F = F_{E}[/tex]
[tex]mg = qE[/tex]
[tex]|q| = \frac{mg}{E}[/tex]
the magnitude of the charge:
[tex]|q| = \frac{0.004\times 9.8g}{700} = 5.6\times 10^{- 5}\ C[/tex]
Since, the nature of the charge is negative:
[tex]q = - 5.6\times 10^{- 5}\ C[/tex]
The electric force per unit charge is defined as the electric field. In this, A charge has a negative sign and a magnitude of [tex]-5.6057 \times 10^{-5} \ C\\\\[/tex].
Electric field direction:
The direction of the electric field[tex]\to[/tex] on the page.
- A force created by an electric field should be in the opposite direction of the electric field.
- As a consequence, the charge should be negative, which is defined in the attached file please find it.
[tex]\to qE = mg[/tex] (for equilibrium)
[tex]\to q\times 700 = 0.00 4 \times 9.81\\\\ \to |q|= 5.6057 \times 10^{-5} \ C\\\\ \to q= -5.6057 \times 10^{-5} \ C\\\\[/tex]
Find out more about the electric field here:
brainly.com/question/8971780
