Respuesta :
Answer:
the sign and magnitude of the charge is - 2 x 10⁻⁵ C.
Explanation:
Given;
mass of the particle, m = 1.43 g = 0.00143 kg
electric field experienced by the particle, E = 700 N/C
The force experienced by the particle is calculated as;
F = mg = EQ
Where;
Q is the magnitude of the charge
[tex]Q = \frac{mg}{E} \\\\Q = \frac{0.00143 \times 9.8}{700} \\\\Q = 2\times 10^{-5} \ C[/tex]
The force must be upward in opposite direction to the electric field. Since the force and the electric field are in opposite direction, the charge must be negative.
Therefore, the sign and magnitude of the charge is - 2 x 10⁻⁵ C.
The particle to remain stationary, when placed in a downward-directed electric the force must be in opposite direction which upward directed.
The charge of the given particle to remain stationary should be [tex]-2\times10^{-5}[/tex] C.
What is electric charge?
The electric force experienced by the body when placed it into the electromagnetic field is called electric charge.
Given information-
The mass of the particle is 1.43 g or 0.00143 kg.
The magnitude of the downward-directed electric field is 700 N/C.
The magnitude of the free-fall acceleration is 9.80 meter per second squared.
The electric field is defined as the electric force per unit charge. It can be given as,
[tex]E=\dfrac{F}{q}[/tex]
Rewrite the equation for the charge,
[tex]q=\dfrac{F}{E}[/tex]
Force experienced by the particle is equal to the product of mass and free fall acceleration (gravity). Thus,
[tex]q=\dfrac{0.00143\times9.8}{700}\\q=2\times10^{-5}[/tex]
Thus the magnitude of the charge is [tex]2\times10^{-5}[/tex] C.
The particle to remain stationary, when placed in a downward-directed electric the force must be in opposite direction which upward directed. For the opposite direction the sign of the charge should be negative.
Thus the charge of the given particle to remain stationary should be [tex]-2\times10^{-5}[/tex] C.
Learn more about the electric charge here;
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