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Answer :
(1) The new voltage across the plates of a parallel-plate capacitor filled with a dielectric is (a) 10 V. (2) The plates of a parallel-plate capacitor connected to a 6V battery is 9.6 x 10⁻¹⁹ J.
1) To determine the new voltage across the plates of a parallel-plate capacitor, we need to consider the effect of the dielectric.
The voltage across the plates of a capacitor is directly proportional to the electric field between the plates.
When the capacitor is connected to a battery and charged to a potential difference of V = 120 V, the electric field is established between the plates.
When the gap between the plates is filled with a dielectric material (k = 12), the dielectric constant affects the electric field.
The electric field decreases by a factor of k when the dielectric is introduced.
In this case, the electric field decreases to 1/12 of its initial value.
Since the voltage across the plates is directly proportional to the electric field, the new voltage across the plates will also decrease by a factor of k, becoming 1/12 of the initial voltage.
Therefore, the new voltage across the plates is V/k = 120 V / 12 = 10 V.
2) The potential energy of an electron placed exactly halfway between the plates of a parallel-plate capacitor can be determined using the formula for the potential energy of a charged particle in an electric field.
The potential energy of a charged particle in an electric field is given by the formula U = qV, where U is the potential energy, q is the charge of the particle, and V is the potential difference.
In this case, the electron has a charge of -e (negative charge) and the potential difference across the plates of the capacitor is 6 V.
Plugging these values into the formula, we have:
U = (-e)(6 V) = -6e V
To calculate the potential energy in joules, we can use the charge of an electron, which is approximately 1.6 x 10^(-19) C.
U = (-1.6 x 10⁻¹⁹ C)(6 V) = -9.6 x 10⁻¹⁹ J
However, the potential energy is a scalar quantity and does not have a negative sign. So, the magnitude of the potential energy is 9.6 x 10⁻¹⁹ J.
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