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sattari [20]
3 years ago
14

Which is described as the flow of charge particles?

Physics
1 answer:
Lady bird [3.3K]3 years ago
7 0
Electric current

For ex. The particles are in a strict
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Two fully charged cylindrical capacitors are connected to two identical batteries. The capacitors are identical except that the
Leni [432]

Answer:

Part(a):  The relative capacitance is \dfrac{C_{A}}{C_{B}} = 0.33

Part(b): The relative energy stored is \dfrac{U_{A}}{U_{B}} = 0.33

Part(c): The relative charge stored is \dfrac{Q_{A}}{Q_{B}} = 0.33

Explanation:

We know the capacitance (C) of a capacitor having charge (Q) and subjected to a potential difference of (V) is given by

C = \dfrac{Q}{V}

Also, the energy (U) stored by a capacitor can be written as

U = \dfrac{1}{2}C~V^{2}

Let us assume that the inner radius of the Capacitor B, as shown in the figure, be \textbf{r_{i}^{B}}\bf{r_{i}^{B}}, the outer radius be \bf{r_{o}^{B}}, the inner radius of Capacitor A be \bf{r_{i}^{A}} and the outer radius be \bf{r_{o}^{B}}.

Given in the problem,

&& r_{o}^{B} = 2~r_{B}^{i}\\&& r_{o}^{A} = 4~r_{B}^{i}\\&& and~r_{i}^{B} = 4~r_{o}^{B} = 8~r_{B}^{i}

Now, the capacitance (C) of a cylindrical capacitor is given by,

\bf{C = \dfrac{2~\pi~\epsilon_{0}~L}{ln(\dfrac{r_{o}}{r_{i}})}}

where \epsilon_{o} is the permittivity of the free space, L is the length of the cylindrical capacitor.

Part(a):

The capacitance of capacitor A,

C_{A} = \dfrac{2~\pi~\epsilon_{0}L}{ln(\dfrac{r_{o}^{A}}{r_{i}^{A}})} = \dfrac{2~\pi~\epsilon_{0}L}{ln(\dfrac{8~r_{i}^{B}}{r_{i}^{B}})} = \dfrac{2~\pi~\epsilon_{0}L}{ln(8)}

and the capacitance of capacitor B,

C_{B} = \dfrac{2~\pi~\epsilon_{0}L}{ln(\dfrac{r_{o}^{B}}{r_{i}^{B}})} = \dfrac{2~\pi~\epsilon_{0}L}{ln(\dfrac{2~r_{i}^{B}}{r_{i}^{B}})} = \dfrac{2~\pi~\epsilon_{0}L}{ln(2)}

giving the relative capacitance of each capacitor to be

\dfrac{C_{A}}{C_{B}} = \dfrac{ln(2)}{ln(8)} = \dfrac{ln(2)}{3~\ln(2)} = \dfrac{1}{3} = 0.33

Part(b):

Energy stored by capacitor A,

U_{A} = \dfrac{1}{2}~C_{A}~V^{2}

Energy stored by capacitor B,

U_{B} = \dfrac{1}{2}~C_{B}~V^{2}

giving the relative energy stored by each capacitor to be

\dfrac{U_{A}}{U_{B}} = \dfrac{C_{A}}{C_{B}} = 0.33

Part(c):

The charge stored by capacitor A,

Q_{A} = C_{A}~V

The charge stored by capacitor B,

Q_{B} = C_{B}~V

giving the relative charge stored by each capacitor to be

\dfrac{Q_{A}}{Q_{B}} =  \dfrac{C_{A}}{C_{B}} = 0.33

8 0
3 years ago
Four charges 7 × 10−9 C at (0 m, 0 m), −9 × 10−9 C at (3 m, 3 m), 7 × 10−9 C at (1 m, 3 m), and −8 × 10−9 C at (−3 m, 2 m), are
Ivanshal [37]

Answer:

Magnitude of the resulting force on the 7 nC charge at the origin:

Fn₁= 23.95*10⁻⁹ N

Explanation:

Look at the attached graphic:

Charges of positive signs exert repulsive forces on q₁ + and charges of negative signs exert attractive forces on q₁ +.

q₁ experiences three forces (F₂₁,F₃₁,F₄₁) and we calculate them with Coulomb's law:

F = (k*q₁*q)/(d)²

d_{12} = \sqrt{3^{2}+3^{2}  }  = \sqrt{18} m : distance from q₁ to q₂

(d₁₂)² = 18 m²

d_{13} =\sqrt{1^{2}+3^{2}  } = \sqrt{10} m  : distance from q₁ to q₃

(d₁₃)² = 10 m²

d_{14} =\sqrt{3^{2}+2^{2}  } = \sqrt{13} m  : distance from q₁ to q₄

(d₁₄)² = 13 m²

K=  8.98755 × 10⁹ N *m²/C²

q₁=  7*10⁻⁹C

k*q₁=8.98755*10⁹ *7*10⁻⁹= 62.9

F₂₁= (62.9)*(9* 10⁻⁹) /(18) = 31.45*10⁻⁹ C

F₃₁= (62.9)*(7* 10⁻⁹) /(10) = 44*10⁻⁹ C

F₄₁= (62.9)*(8* 10⁻⁹) /(13) = 38.7*10⁻⁹ C

x-y components of the net force on q₁ (Fn₁):

α= tan⁻¹(3/3)= 45°  ,  β= tan⁻¹(3/1)= 71.56° , θ= tan⁻¹(2/3)= 33.69°

Fn₁x = F₂₁x+ F₃₁x+F₄₁x

F₂₁x =+ F₂₁*cosα =+ (31.45*10⁻⁹)* (cos 45°) = +22.24 *10⁻⁹ N

F₃₁x= -F₃₁*cosβ = - ( 44*10⁻⁹)* (cos 71.56°) = -13.91 *10⁻⁹ N

F₄₁x= -F₄₁*cosθ = -(38.7*10⁻⁹)* (cos 33.69°) = -32.2*10⁻⁹ N

Fn₁x = (+22.24 - 13.91 - 32.2)*10⁻⁹ N

Fn₁x = -23.87 *10⁻⁹ N

Fn₁y = F₂₁y+ F₃₁y+F₄₁y

F₂₁x =+ F₂₁*sinα =+ (31.45*10⁻⁹)* (sin 45°) = +22.24 *10⁻⁹ N

F₃₁x= -F₃₁*sinβ = - ( 44*10⁻⁹)* (sin 71.56°) = -41.74 *10⁻⁹ N

F₄₁x= +F₄₁*sinθ = +(38.7*10⁻⁹)* (sin 33.69°) =+21.47*10⁻⁹ N

Fn₁y = (22.24 -41.74+21.47)*10⁻⁹ N  

Fn₁y = 1.97*10⁻⁹ N

Magnitude of the resulting force on the 7 nC charge at the origin (q₁):

F_{n1} =\sqrt{(Fn_{1x} )^{2}+(Fn_{1y} )^{2} }

F_{n1} =\sqrt{(23.87 )^{2}+(1.97 )^{2} }

Fn₁= 23.95*10⁻⁹ N

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