Answered

What must the charge (sign and magnitude) of a particle of mass 1.41 gg be for it to remain stationary when placed in a downward-directed electric field of magnitude 670 N/CN/C ?

Answer :

Answer:

[tex]q = 2.067 \times 10^{-5}\ C[/tex]

Explanation:

Given,

mass = 1.41 g = 0.00141 Kg

Electric field,E = 670 N/C.

We know,

Force in charge due to Electric field.

F = E q

And also we know

F = m g

Equating both the equation of motion

m g = E q

[tex]q =\dfrac{mg}{E}[/tex]

[tex]q =\dfrac{0.00141 \times 9.81}{670}[/tex]

[tex]q = 2.067 \times 10^{-5}\ C[/tex]

Charge of the particle is equal to [tex]q = 2.067 \times 10^{-5}\ C[/tex]

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