Respuesta :
Answer:
a) The charge of the capacitor is 4.25x10⁻¹¹C
b) The charge of the capacitor is 4.25x10⁻¹¹C because the battery is disconnected.
c) The potential difference across the plates is 18 V
d) The work is 7.64x10⁻¹⁰J
Explanation:
The capacitance of the capacitor is equal to:
[tex]C=\frac{e_{0}A }{d}[/tex]
A = 2 cm² = 0.0002 m²
d = 0.5 mm = 0.0005 m
Replacing:
[tex]C=\frac{8.85x10^{-12}*0.0002 }{0.0005} =3.54x10^{-12} F = 3.54pF[/tex]
a) The charge of the capacitor is equal to:
Q = C*V = 3.54 * 12 = 42.48 pC = 4.25x10⁻¹¹C
b) The charge is the same because the battery is disconnected (Q = 4.25x10⁻¹¹C)
c) If distance is increased, we have:
[tex]C=\frac{8.85x10^{-12}*0.0002 }{0.00075} =2.36x10^{-12} F=2.36pF[/tex]
The potential is:
[tex]V=\frac{Q}{C} =\frac{42.48}{2.36} =18V[/tex]
d) The work done is equal to:
[tex]W=VQ=18*42.48=764.6pJ=7.64x10^{-10} J[/tex]
Answer:
(a) The charge on the capacitor, Q = 42.5 pC
b) The new charge on the capacitor, Q = 42.5 pC
c) New voltage across the capacitor, V = 18.0 V
d) Work done, W = 127.5 pJ
Explanation:
The capacitance of a parallel plate capacitor is given by the relation:
[tex]C = \frac{k \epsilon_{0}A }{d}[/tex]..................(1)
Area of plate, [tex]A = 2 cm^{2} = 2 * 10^{-4} m^{2}[/tex]
For air, dielectric constant k = 1
Initial potential difference, V₀ = 12.0 V
Separation between the plates, d = 0.50 mm = 0.50 x 10⁻³ m
The charge on the capacitor can be given by the equation
[tex]Q = C V_{0}[/tex]...............(2)
Putting appropriate values into equation (1)
[tex]C = \frac{1 * 8.854 * 10^{-12}* 2 * 10^{-4} }{0.5 * 10^{-3} }[/tex]
[tex]C = 3.542 * 10^{-12} F[/tex]
Inserting the value of C into equation (2)
[tex]Q = 3.542 * 10^{-12} * 12\\Q = 42.5 * 10^{-12} C\\Q = 42.5 pC[/tex]
(b) Since the battery has been disconnected from the capacitor, the charge on the capacitor does not change despite the increase in the separation between the plates
[tex]Q = 42.5 * 10^{-12} C[/tex]
(c) New potential difference between the plates,
Since the charge remains the same after the disconnection
[tex]Q_{f} = Q_{i}[/tex]
[tex]C_{f} V_{f} = C_{i} V_{i}[/tex]
[tex]\frac{k \epsilon_{0}A }{d_{f} }V_{f} = \frac{k \epsilon_{0}A }{d_{i} }V_{i}[/tex]
[tex]\frac{V_{f} }{d_{f} } = \frac{V_{i} }{d_{i} } \\\frac{V_{f} }{0.75 * 10^{-3} } = \frac{12 }{0.5 * 10^{-3} } \\V_{f} = \frac{12 *0.75 * 10^{-3} }{0.5 * 10^{-3}} \\V_{f} = 18.0 V[/tex]
The new potential difference across the plates is 18 V
(d) Amount of Work required to pull the plates to their new separation -
[tex]W = \frac{1}{2} q(V_{2} - V_{1} )\\W = \frac{1}{2} * 42.5 * 10^{-12} (18-12 )\\W = 0.5 * 42.5 * 6 * 10^{-12}\\W = 127.5 * 10^{-12} J\\W = 127.5 pJ[/tex]
Work done = 127.5 pJ
