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dimulka [17.4K]
3 years ago
7

A spherical capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has a radi

us of ra = 12.3 cm , and the outer sphere has a radius of rb = 15.1 cm . A potential difference of 120 V is applied to the capacitor.
(a) What is the capacitance of the capacitor?
(b) What is the magnitude E1 of the electric field E? at radius r= 12.8cm, just outside the inner sphere?
(c) What is the magnitude of E? at r= 14.7cm, just inside the outer sphere?
Physics
1 answer:
Svetlanka [38]3 years ago
8 0

Answer: a) 73.41 10^-12 F; b)4.83* 10^3 N/C; c) 3.66 *10^3 N/C

Explanation: To solve this problem we have to consider the following: The Capacity= Charge/Potential Difference

As we know the capacity is  value that depend on the geometry of the capacitor, in our case two concentric spheres.

So Potential Difference between the spheres is given by:

ΔV=-\int\limits^a_b {E} \, dx

Where E = k*Q/ r^2

so we have \int\limits^a_b {K+Q1/r} \, dr

then

Vb-Va=k*Q(1/b-1/a)=kQ (ab/b-a)

Finally using C=Q/ΔV=ab/(k(b-a))

To caclulate the electric firld we first obtain the charge

Q=ΔV*C=120 V*73.41 10^-12 F=8.8 10^-9 C

so E=KQ/r^2 for both values of r

r=12.8 cm ( in meters)

r2=14.7 cm

E(r1)=4.83* 10^3 N/C

E(r2)=3.66 *10^3 N/C

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A cell has an internal resistance of 0.02ohms and e.m.f of 2.0v calculate it's terminal p.d if it's delivers (a)5A ( b)50A​
Aleksandr-060686 [28]

Answer:

(a) The terminal voltage of the cell is <u>1.9 V.</u>

(b) The terminal voltage of the cell is <u>1.0 V.</u>

Explanation:

Given:

(a)

E.M.F of the cell (E) = 2.0 V

Internal resistance of the cell (r) = 0.02 ohms

Current passing through the cell (I) = 5 A

Now, the potential difference across the terminals of the cell is given as:

V=E-Ir

Plug in the given values and solve for 'V'. This gives,

V=2.0-(5\times 0.02)\\\\V=2.0-0.1=1.9\ V

Therefore, the terminal voltage of the cell is 1.9 V.

(b)

E.M.F of the cell (E) = 2.0 V

Internal resistance of the cell (r) = 0.02 ohms

Current passing through the cell (I) = 50 A

Now, the potential difference across the terminals of the cell is given as:

V=E-Ir

Plug in the given values and solve for 'V'. This gives,

V=2.0-(50\times 0.02)\\\\V=2.0-1.0=1.0\ V

Therefore, the terminal voltage of the cell is 1.0 V.

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1 Amp = 1 Coulomb/sec
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In 10 second, number of electrons are calculated as;
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