What are the sign and magnitude in coulombs of a point charge that produces a potential of -3.50 V at a distance of 2.00 mm

Physics

1 answer:

DanielleElmas [232]2 years ago

5 0

The magnitude and sign of the charge are 0.8 MC and negative respectively.

To find the answer, we need to know about the electric potential of a point charge.

<h3>What's the mathematical expression of the electric potential of a point charge?</h3>

Mathematically, the electric potential at a distance 'r' from a point charge 'q' is given as (Kq)/r.
K is the electrical constant with value 9×10^(-9) in vaccum.

<h3>What will be the magnitude and sign of a charge, if potential is -3.50V at 2mm?</h3>

From the expression of electric potential, charge is

q= (potential ×r)/K

= (-3.5×0.002)/ (9×10^(-9))

= -0.8 mega coulomb.

Thus, we can conclude that the magnitude and sign of the charge are 0.8 MC and negative respectively.

Learn more about the electric potential here:

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The near point of a person's eye is 71.4 cm. To see objects clearly at a distance of 24.0 cm, what should be the focal length an

yuradex [85]

Answer:

The focal length of the appropriate corrective lens is 35.71 cm.

The power of the appropriate corrective lens is 0.028 D.

Explanation:

The expression for the lens formula is as follows;

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

Here, f is the focal length, u is the object distance and v is the image distance.

It is given in the problem that the given lens is corrective lens. Then, it will form an upright and virtual image at the near point of person's eye. The near point of a person's eye is 71.4 cm. To see objects clearly at a distance of 24.0 cm, the corrective lens is used.

Put v= -71.4 cm and u= 24.0 cm in the above expression.

$\frac{1}{f}=\frac{1}{24}+\frac{1}{-71.4}$