Department of Pre-University Education, Karnataka course PUC Karnataka Science Class 12
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# 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 - Physics

ConceptElectric Potential Difference

#### 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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Solution 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 Concept: Electric Potential Difference.
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