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A charge of −−3.00 nC is placed at the origin of an xyxy-coordinate system, and a charge of 2.00 nC is placed on the yy-axis at y=y= 4.00 cm. (a) If a third charge, of 5.00 nC, is now placed at the point x=x= 3.00 cm, y=y= 4.00 cm, find the xx- and yy-components of the total force exerted on this charge by the other two charges. (b) Find the magnitude and direction of this force.

Respuesta :

Answer:

Explanation:

a )

Force on 5 nC due to -3 nC

= 9 x 10⁹x 5 x 3 x 10⁻¹⁸ / ( 5 x 10⁻²)²

= 5.4 x 10⁻⁵ N

X component = 5.4 x 10⁻⁵ cosθ

= -5.4 x 10⁻⁵ x 3/5

= -3.24 x 10⁻⁵ N

y component

= -5.4 x 10⁻⁵ x 4/5

= -4.32 x  10⁻⁵ N

Force on 5 nC due to 2 nC

= 9 x 10⁹x 5 x 2 x 10⁻¹⁸ / ( 3 x 10⁻²)²

= 10 x 10⁻⁵ N

x component = 10 x 10⁻⁵ N

Total x component = (10-3.24)x10⁻⁵ N

= 6.76 x 10⁻⁵ N

y component =   -4.32 x  10⁻⁵ N

magnitude = √( 6.76² + 4.32²) x 10⁻⁵ N

= 8 .02 N

DIRECTION =

Tan θ = -4.32 / 6.76

θ = 32.5

with negative x -direction , south west.

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