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Fiesta28 [93]
3 years ago
15

An object is 2.0 cm from a double convex lens with a focal length of 1.5 cm. Calculate the image distance

Physics
1 answer:
Tomtit [17]3 years ago
3 0

Answer:

0.857 cm

Explanation:

We are given that:

The focal length for a convex lens to be (f) = 1.5cm

The object distance (u) = - 2.0 cm

We are to determine the image distance (v) = ??? cm

By applying the lens formula:

\dfrac{1}{f} = \dfrac{1}{u}+\dfrac{1}{v}

By rearrangement and making (v) the subject of the above formula:

v = \dfrac{uf}{u-f}

replacing the given values:

v = \dfrac{(-2.0)(1.5)}{(-2.0 -1.5)}

v = \dfrac{-3.0}{(-3.5)}

v = 0.857 cm

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Estimate the electric field at a point 2.40 cm perpendicular to the midpoint of a uniformly charged 2.00-m-long thin wire carryi
nadya68 [22]

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E = 1.85*10^{12}\frac{N}{C}

Explanation:

Hi!

The perpendicular distance 2.4cm, is much less than the distance to both endpoints of the wire, which is aprox 1m. Then the edge effect is negligible at this field point, and we can aproximate the wire as infinitely long.

The electric filed of an infinitely long wire is easy to calculate. Let's call z the axis along the wire. Because of its simmetry (translational and rotational), the electric field E must point in the radial direction,  and it cannot depende on coordinate z. To calculate the field Gauss law is used, as seen in the image, with a cylindrical gaussian surface. The result is:

E = \frac{\lambda}{2\pi \epsilon_0 r}\\\lambda=\text{charge per unit length}=\frac{4.95 \mu C}{2 m} = 2.475 \frac{C}{m}\\r=\text{perpendicular distance to wire}\\\epsilon_0=8.85*10^{-12}\frac{C^2}{Nm^2}

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