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What must the charge (sign and magnitude) of a particle of mass 1.48 g be for it to remain stationary when placed in a downward-directed electric field of magnitude 640 N/C

Sagot :

Answer:

[tex]q=-2.26\times 10^{-5}\ C[/tex]

Explanation:

Given that,

The mass of a particle, m = 1.48 g = 0.00148 kg

The electric field, E = 640 N/C

We need to find the charge of the particle when placed in a downward-directed electric field.

The force of gravity is balanced by the electric force such that,

mg = qE

Where

q is the charge of the particle

[tex]q=\dfrac{mg}{E}\\\\q=\dfrac{0.00148\times 9.8}{640}\\\\q=2.26\times 10^{-5}\ C[/tex]

q must be negative, the force must be upward (opposite direction of the electric field).

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