A parallel plate capacitor is charged from a cell and then isolated fr...
Explanation:
When a parallel plate capacitor is charged from a cell and then isolated from it, several changes occur. Let's examine each option individually:
A) The force of attraction between the plates will decrease:
The force of attraction between the plates of a capacitor is given by the equation F = (Q^2) / (2εA), where Q is the charge on the plates, ε is the permittivity of the medium between the plates, and A is the area of the plates. The force of attraction is directly proportional to the charge on the plates. Therefore, when the separation between the plates is increased, the charge on the plates remains constant, and the force of attraction between the plates decreases. So, option A is correct.
B) The energy stored in the capacitor will increase:
The energy stored in a capacitor is given by the equation U = (1/2) (Q^2) / C, where U is the energy stored, Q is the charge on the plates, and C is the capacitance of the capacitor. The energy stored is directly proportional to the square of the charge on the plates. When the separation between the plates is increased, the capacitance of the capacitor decreases according to the equation C = (εA) / d, where ε is the permittivity of the medium between the plates and d is the separation between the plates. As the capacitance decreases, the energy stored in the capacitor increases. So, option B is correct.
C) The potential difference between the plates will decrease:
The potential difference between the plates of a capacitor is given by the equation V = Q / C, where V is the potential difference, Q is the charge on the plates, and C is the capacitance of the capacitor. When the separation between the plates is increased, the capacitance of the capacitor decreases, as mentioned earlier. As a result, the potential difference between the plates also decreases according to the equation V = Q / C. So, option C is correct.
D) The electric field in the region between the plates will not change:
The electric field between the plates of a capacitor is given by the equation E = V / d, where E is the electric field, V is the potential difference between the plates, and d is the separation between the plates. When the separation between the plates is increased, the potential difference between the plates decreases, but the separation also increases proportionally. As a result, the electric field between the plates remains constant. So, option D is correct.
In conclusion, the correct options are B and D: The energy stored in the capacitor will increase, and the electric field in the region between the plates will not change.
A parallel plate capacitor is charged from a cell and then isolated fr...
The correct answers are: The field in the region between the plates will not change, The energy stored in the capacitor will increase