Q: 11 (CDS-I/2017)
question_subject:
Science
question_exam:
CDS-I
stats:
0,1,8,1,4,1,3
keywords:
{'capacitance': [0, 0, 0, 1], 'plate capacitor': [0, 0, 0, 1], 'capacitor': [0, 0, 1, 3], 'dielectric': [0, 0, 0, 1], 'plates': [0, 0, 1, 4], '1c': [0, 0, 0, 1]}
When the space between the two plates of a parallel-plate capacitor, initially filled with air, is replaced with a dielectric material, the capacitance of the capacitor increases.
The value of the capacitance will become equal to the initial capacitance (C) multiplied by the dielectric constant (k) of the dielectric material. In this case, the dielectric constant is given as 7.
Therefore, the value of the capacitance will become 7C, which is option 3.
It is important to note that the presence of a dielectric material between the plates of a capacitor increases the capacitance because the dielectric material reduces the electric field between the plates, effectively storing more charge for the same potential difference.