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AysviL [449]
3 years ago
11

A parallel plate capacitor is charged up by a battery. The battery is then disconnected, but the charge remains on the plates. T

he plates are then pulled apart. Explain whether each of the following quantities increases, decreases, or remains the same as the distance between the plates increases for each of these cases
.
(a) the capacitance of the capacitor(b) the potential difference between the plates(c) the electric field between the plates(d) the electric potential energy stored by the capacitor
Physics
1 answer:
larisa [96]3 years ago
7 0

Answer:

a. Decreases

b. Increases

c. Remains the same

d. Increases

Explanation:

a. Capacitance is given by c= Ak€/d

where A is conductivity plate with Area

K is a constant

€ is dielectric with permittivity.

d is the distance

b. Potential difference is given by

V = Ed, since, the electric field remains the

same, the potential diterence also increases with increase in distance.

Since the capacitance depends upon the distance, and all the other factors are kept constant, the capacitance decreases.

c. Electric field remains the same because charge on the

plate remains the same.

d. since electric field remains the same and capacitance decreases, the energy increases.

E= 1/2c * Q^2

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Question:

A point charge of -2.14uC  is located in the center of a spherical cavity of radius 6.55cm  inside an insulating spherical charged solid. The charge density in the solid is 7.35×10−4 C/m^3.

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Express your answer using two significant figures.

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The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity 3.65\times 10^5N/C

Explanation:

A point charge ,q = -2.14\times 10^{-6} C is located in the center of a spherical cavity of radius , r =6.55\times 10^{-2}  m inside an insulating spherical charged solid.  

The charge density in the solid , d = 7.35 \times 10^{-4}C/m^3.

Distance from the center of the cavity,R =9.5\times 10^{-2 }m

Volume of shell of charge= V  =(\frac{4\pi}{3})[ R^3 - r^3 ]

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Q = 4.1888*\times 10^{-4 }[8.57375 - 2.81011 ]\times 7.35\times 10^{-4}

Q = 4.1888\times 10^{-4} [5.76364 ] \times 7.35 \times 10^{-4}

Q =2.4143 \times 10^{-4} \times 7.35 \times 10^ { -4}

Q =1.7745 \times 10^{-6 }C

Electric field at 9.5\times 10^{-2}m due to shellE1  = \frac{k Q}{R^2}

E1 =  \frac{ 9 \times 10^9\times 1.7745\times 10^{-6 }}{ 90.25\times 10^{-4}}

E1 =1.769\times 10^6 N/C

Electric field at  9.5\times 10^{-2} due to 'q' at center E2 = \frac{kq}{R^2}

E2 =\frac{ - 9 \times 10^9\times 2.14\times 10^{-6 }}{ 90.25\times 10^{-4}}

E2 =2.134\times 10^6 N/C

The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity

= E2- E1

=[  2.134  - 1.769 ]\times 10^6

= 3.65\times 10^5 N/C

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