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Question
A capacitor having a capacitance of 100 µF is charged to a potential difference of 50 V. (a) What is the magnitude of the charge on each plate? (b) The charging battery is disconnected and a dielectric of dielectric constant 2⋅5 is inserted. Calculate the new potential difference between the plates. (c) What charge would have produced this potential difference in absence of the dielectric slab. (d) Find the charge induced at a surface of the dielectric slab.
Solution
(a) The magnitude of the charge can be calculated as:
Charge = Capacitance × Potential difference
⇒ `Q = 100 xx 10^-6 xx 50 = 5 "mC"`
(b) When a dielectric is introduced, the potential difference decreases.
We know,
`V = "Initial potential"/"Dielectric Constant"`
⇒ `V = 50/2.5 = 20V`
(c) Now, the charge on the capacitance can be calculated as:
Charge = Capacitance × Potential difference
⇒ `q_f = 20 xx 100 xx 10^-6 = 2 "mC"`
(d) The charge induced on the dielectric can be calculated as :
`q = q_i(1-1/K) = 5 "mC" (1 - 1/2.5) = 3 "mC"`
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