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3 years ago

15

What must the charge (sign and magnitude) of a particle of mass 1.49 g be for it to remain stationary when placed in a downward-

directed electric field of magnitude 670 n/c ?

1 answer:

Kipish [7]3 years ago

6 0

Answer;

= -2.18 × 10^-5 C

Explanation;

m = 1.49 × 10^-3 kg

Take downward direction as positive.

Fg = m g

E = 670 N/C

Fe = q E

Fe + Fg = 0

q E + m g = 0

q = -m g/E

= -1.49 × 10^-3 × 9.81/670

= -2.18 × 10^-5 C

= -2.18 × 10^-5 C

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Explanation:

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We need to find the angle $\theta$ at which Ferdinand jumps.

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Substituting (3) in (4):

$4 m/s=5 m/s cos\theta$ (5)

Clearing $\theta$ :

$\theta=cos^{-1} (\frac{4 m/s}{5 m/s})$

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3 years ago

Read 2 more answers