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IrinaVladis [17]
3 years ago
6

A straight, non-conducting plastic wire 8.50 cm long carries a charge density of 175 nC/m distributed uniformly along its length

. It is lying on a horizontal tabletop. a) Find the magnitude and direction of
Physics
1 answer:
Marina86 [1]3 years ago
3 0

Answer:

\mathbf{E = 3.04 \times 10^4 \ N/C} and the direction is vertically upward

Explanation:

To find the magnitude & direction which was being produced at a given point of 6.00 cm which is directly above the midpoint.

Given that:

the wire length = 8.50 cm = 8.5 × 10⁻² m

the charge density λ = 175 n C/m = 175 ×10⁻⁹ C/m

distance x = 6.00 cm = 0.006 m

Assume that  N be the point at distance x situated directly above the midpoint

Then, the linear expression for the linear charge density is:

\lambda = \dfrac{Q}{L}

the charge Q = \lambda L

Q = 175 \times 10^{-9} \times 8. 5 \times 10^{-2}

Q = 1.49 \times 10^{-8} \ C

For the electric field at point N;

E = \dfrac {kQ}{x\sqrt{x^2 + r^2}}

where;

r = \dfrac{L} {2}

E = \dfrac {9\times 10^9 \times 1.49 \time 10^{-8} }{0.06 \times \sqrt{0.06^2 + (\dfrac{8.50 \times 10^{-2}}{2})^2}}

\mathbf{E = 3.04 \times 10^4 \ N/C}

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Tju [1.3M]

The electric field strength will be 0.6252 V/m. It is the strength at which the field is created by charges.

<h3>What is electric file strength?</h3>

The electric field strength is defined as the ratio of electric force and charge.

The electric field strength is found as;

\rm E = \frac{I \rho }{A} \\\\ \rm E = \frac{20  \times 1.68 \times 10^{-8} }{ (0.6385 \times 10^{-6}} \\\\ E= 0.5262 \  V/m

Hence, the electric field strength will be 0.6252 V/m.

To learn more about the electric field strength, refer to the link;

brainly.com/question/4264413

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a

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c

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What is the kinetic energy of an object that has a mass of 50.0 kg and a velocity of 18 m/s?
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k=1/2mv^2


k=1/2 x 50.0 x 18^2 = 8100 joules

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4 years ago
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