Electrostatics Test

Result

Electrostatics Test - 62

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Weekly Quiz Competition

Question 1
4 / -1

A parallel plate capacitor is first charged and then a dielectric slab is introduced between the plates. The quantity that remains unchanged is

A
Charge Q

B
Potential V

C
Capacity C

D
Energy U
Question 2
4 / -1

The force between the plates of a parallel plate capacitor of capacitance C and distance of separation of the plates d with a potential difference V between the plates, is

A

B

C

D
Question 3
4 / -1

Two metal spheres of capacitance C₁ and C₂ carry some charges. They are put in contact and then separated. The final charges Q₁ and Q₂ on them will satisfy

A

B

C

D
Question 4
4 / -1

What is the area of the plates of a 3F parallel plate capacitor, if the separation between the plates is 5 mm

A
1.694 × 10⁹ m²

B
4.529 × 10⁹ m²

C
9.281 × 10⁹ m²

D
12.981 × 10⁹ m²

Solution
Question 5
4 / -1

A parallel plate capacitor has circular plates of 0.08 m radius and 1.0 × 10^–3 m separation. If a P.D. of 100 volt is applied, the charge will be

A
1.8 × 10^–10 C

B
1.8 × 10^–8 C

C
1.8 × 10^–20 C

D
None of these

Solution
Question 6
4 / -1

The capacity of a parallel plate condenser is 10 µF without dielectric. Dielectric of constant 2 is used to fill half the distance between the plates, the new capacitance in µF is

A
10

B
20

C
15

D
13.33

Solution
Question 7
4 / -1

A battery is used to charge a parallel plate capacitor till the potential difference between the plates becomes equal to the electromotive force of the battery. The ratio of the energy stored in the capacitor and the work done the battery will be

A
1

B
2

C
1/4

D
1/2

Solution
Question 8
4 / -1

Two protons A and B are placed in space between plates of a parallel plate capacitor charged upto V volts (See fig.). Forces on protons are F_A and F_B, then

A
F_A > F_B

B
F_A < F_B

C
F_A = F_B

D
Nothing can be said

Solution
Question 9
4 / -1

A parallel plate capacitor has a plate separation of 0.01 mm and use a dielectric (whose dielectric strength is 19KV / mm) as an insulator. The maximum potential difference that can be applied to the terminals of the capacitor is

A
190 V

B
290 V

C
95 V

D
350 V

Solution
Question 10
4 / -1

A variable condenser is permanently connected to a 100V battery. If the capacity is changed from 2 µF to 10 µF, then change in energy is equal to

A
2 × 10^–2 J

B
2.5 × 10^–2 J

C
3.5 × 10^–2 J

D
4 × 10^–2 J

Solution