BITSAT 2024 | Capacitor Question 1 | Physics

A parallel plate capacitor consists of two circular plates of radius $ R=0.1 \mathrm{~m} $. They are separated by a short distance. If electric field between the capacitor plates changes as $ \frac{d E}{d t}=6 \times 10^{13} \frac{\mathrm{~V}}{\mathrm{~m} \times \mathrm{s}} $, then the value of displacement current is

15.25 A

6.25 A

16.67 A

4.69 A

BITSAT 2021

MCQ (Single Correct Answer)

-1

The separation between two parallel plates of capacitor is 1 mm. What is the electric potential generates between the plates of capacitor, when electric field of 2000 N/C is applied on it?

2 V

2000 V

0.2 V

200 V

BITSAT 2020

MCQ (Single Correct Answer)

-1

Three capacitors C₁, C₂ and C₃ are connected as shown in the figure below. If capacitor C₃ breaks down electrically, the change in total charge on the combination of capacitors is

BITSAT 2020 Physics - Capacitor Question 2 English

$$({C_1} + {C_2})V\left[ {1 - \left( {{{{C_3}} \over {{C_1} + {C_2} + {C_3}}}} \right)} \right]$$

$$({C_3} + {C_2})V\left[ {1 - \left( {{{{C_1}} \over {{C_1} + {C_2} + {C_3}}}} \right)} \right]$$

$$({C_1} + {C_2})V\left[ {1 - \left( {{{{C_2} + {C_1}} \over {{C_1} + {C_3}}}} \right)} \right]$$

$$({C_1} + {C_2})V\left[ {1 - \left( {{{{C_2}} \over {{C_1} + {C_3}}}} \right)} \right]$$

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