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Georgia [21]
3 years ago
15

A parallel-plate capacitor is charged by connecting it across the terminals of a battery. If the battery remains connected and t

he separation between the plates is increased, what will happen to the charge on the capacitor and the voltage across it?
Physics
1 answer:
Tanya [424]3 years ago
6 0

Answer:

The voltage ( or potential difference) V increases while the charge Q decreases.

Explanation:

The capacitance C of a capacitor is defined as the measure to which the capacitor can store charges. For a parallel-plate capacitor it is given by the following relationship;

C=\frac{\epsilon_o\epsilon_rA}{d}............(1)

where A is the surface area of the plates, d is their distance of separation, \epsilon_o is the permittivity of free space and \epsilon_r is relative permittivity.

Also, the capacitance of a capacitor can be expressed in the form of equation (2)

C=\frac{Q}{V}...................(2)\\

where Q is the charge stored and V is the potential difference.

By combining (1) and (2) and making d the subject of formula, we obtain the following;

d=\frac{\epsilon_o\epsilon_rAV}{Q}............(3)

By observing (3), it is seen that the distance d of separation between the plates is directly proportional to the potential difference V and inversely proportional to the charge stored Q. This implies that an increase in the distance d of separation will bring about an increase the the voltage or potential difference V and a decrease in the charge Q.

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Complete Question

The distance between the objective and eyepiece lenses in a microscope is 19 cm . The objective lens has a focal length of 5.5 mm .

What eyepiece focal length will give the microscope an overall angular magnification of 300?

Answer:

The  eyepiece focal length is  f_e  = 0.027 \ m

Explanation:

From the question we are told that

    The focal length is  f_o =  5.5 \ mm =  -0.0055 \ m

This negative sign shows the the microscope is diverging light

     The  angular magnification is m = 300

     The  distance between the objective and the eyepieces lenses is  Z =  19 \ cm  = 0.19 \ m

Generally the magnification is mathematically represented as

        m  =  [\frac{Z - f_e }{f_e}] [\frac{0.25}{f_0} ]

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         f_e  = \frac{Z}{1 - [\frac{M  *  f_o }{0.25}] }

substituting values

        f_e  = \frac{ 0.19 }{1 - [\frac{300  *  -0.0055 }{0.25}] }

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