Respuesta :
Answer:
(a) The charge on capacitor in first case is [tex]8.5 \times 10^{-11}[/tex] C
(b) In second case is [tex]4.25 \times 10^{-11}[/tex] C
(c) Potential across plate is 24 V
(d) Work required to pull the plates is [tex]5.1 \times 10^{-10}[/tex] J
Explanation:
Given:
Area of plate [tex]A = 4 \times 10^{-4}[/tex] [tex]m^{2}[/tex]
Voltage [tex]V = 12[/tex] V
Separation between two plate [tex]d = 0.50 \times 10^{-3}[/tex] m
(a)
The charge on capacitor is given by,
[tex]C = \frac{Q}{V}[/tex]
But capacitance of parallel plate capacitor is given by,
[tex]C = \frac{\epsilon_{o} A }{d}[/tex]
⇒ [tex]Q = \frac{\epsilon _{o}A V }{d}[/tex]
Where [tex]\epsilon _{o} = 8.85 \times 10^{-12}[/tex]
[tex]Q = \frac{8.85\times 10^{-12} \times 4 \times 10^{-4} \times 12}{0.50 \times 10^{-3} }[/tex]
[tex]Q = 8.49 \times 10^{-11}[/tex] C
(b)
The charge on the capacitor when plate separation is [tex]1 \times 10^{-3}[/tex] m is,
Here [tex]d = 1 \times 10^{-3}[/tex] m
[tex]Q = \frac{\epsilon _{o}A V }{d}[/tex]
[tex]Q = \frac{8.85\times 10^{-12} \times 4 \times 10^{-4} \times 12}{1 \times 10^{-3} }[/tex]
[tex]Q = 4.25 \times 10^{-11}[/tex] C
(c)
The potential difference across plate is,
[tex]V = \frac{Q}{C}[/tex]
But [tex]C = \frac{\epsilon_{o} A }{d}[/tex]
[tex]C = \frac{8.85 \times 10^{-12} \times 4 \times 10^{-4} }{1 \times 10^{-3} }[/tex]
[tex]C = 3.45 \times 10^{-12}[/tex] F
Put the value of capacitance and find potential difference,
[tex]V = \frac{8.5 \times 10^{-11} }{3.45 \times 10^{-12} }[/tex]
[tex]V = 24[/tex] V
(d)
Work required is given by,
[tex]W= U_{f} - U_{i}[/tex]
[tex]W = \frac{Q (V_{f} -V_{i} )}{2}[/tex]
Where [tex]V_{f} = 24[/tex] V and [tex]V_{i} = 12[/tex]
[tex]W = \frac{8.5 \times 10^{-11} \times 12}{2}[/tex]
[tex]W = 5.1 \times 10^{-10}[/tex] J
Therefore, the charge on capacitor in first case is [tex]8.5 \times 10^{-11}[/tex] C and in second case is [tex]4.25 \times 10^{-11}[/tex] C and potential across plate is 24 V and work required to pull the plates is [tex]5.1 \times 10^{-10}[/tex] J
(a) Charge will be "[tex]8.49\times 10^{-11} \ C[/tex]".
(b) Charge will be "[tex]4.25\times 10^{-11} \ C[/tex]".
(c) Potential diff. be "[tex]24 \ V[/tex]".
(d) Required work done "[tex]5.1\times 10^{-10} \ J[/tex]".
According to the question,
- Area of plate, [tex]A = 4\times 10^{-4} \ m^2[/tex]
- Voltage, [tex]V = 12 \ V[/tex]
- Distance, [tex]d = 0.50\times 10^{-3} \ m[/tex]
(a)
We know,
→ Charge, [tex]C = \frac{Q}{V}[/tex]
or,
→ [tex]Q = \frac{\epsilon_0 AV}{d}[/tex]
By substituting the values,
[tex]= \frac{8.85\times 10^{-12}\times 4\times 10^{-4}\times 12}{0.50\times 10^{-3}}[/tex]
[tex]= 8.49\times 10^{-11} \ C[/tex]
(b)
The charge will be:
→ [tex]Q = \frac{\epsilon_0 AV}{d}[/tex]
[tex]= \frac{8.85\times 10^{-12}\times 4\times 10^{-4}\times 12}{1\times 10^{-3}}[/tex]
[tex]= 4.25\times 10^{-11} \ C[/tex]
(c)
The potential difference,
→ [tex]V = \frac{Q}{C}[/tex]
or,
→ [tex]C = \frac{\epsilon_0 A}{d}[/tex]
By substituting the values,
[tex]= \frac{8.85\times 10^{-12}\times 4\times 10^{-4}}{1\times 10^{-3}}[/tex]
[tex]= 3.45\times 10^{-12} \ F[/tex]
By substituting the value of "C",
→ [tex]V = \frac{8.5\times 10^{-11}}{3.45\times 10^{-12}}[/tex]
[tex]= 24 \ V[/tex]
(d)
We know the work done,
→ [tex]W = U_f -U_i[/tex]
[tex]= \frac{Q(V_f-V_i)}{2}[/tex]
By substituting the values,
[tex]= \frac{8.5\times 10^{-11}\times 12}{2}[/tex]
[tex]= 5.1\times 10^{-10} \ J[/tex]
Thus the answers above are correct.
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