Electrostatic Potential and Capacitance Test

Result

Electrostatic Potential and Capacitance Test - 6

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Weekly Quiz Competition

Question 1
4 / -1

In a parallel plate capacitor, the capacity increases if

A
area of the plate is decreased

B
distance between the plates increases

C
area of the plate is increased

D
dielectric constant decreases

Solution

In a parallel plate capacitor the capacity of capacitor.

The capacity of capacitor increases if area of the plate is increased.
Question 2
4 / -1

A parallel plate capacitor has two square plates with equal and opposite charges. The surface charge densities on the plate are +σ and −σ respectively. In the region between the plates the magnitude of electric field is:

A
σ/2ε₀

B
σ/ε₀

C
0

D
none of these

Solution

The magnitude of the electric field between the plates is F = σ/2ε₀ - (- σ/2ε₀) - σ/ε₀
Question 3
4 / -1

A parallel plate air capacitor is charged to a potential difference of V volts. After disconnecting the charging battery the distance between the plates of the capacitor is increased using an insulating handle. As a result the potential difference between the plates

A
increases

B
decreases

C
does not change

D
becomes zero

Solution

Capacitance of a parallel plate capacitor
C = ε₀A/d ...(i)
Also capacitance = potential difference/charge .......(ii)
When battery is disconnected and the distance between the plates of the capacitor is increased then capacitance increases and charge remains constant.
Since capacitance = potential difference/charge
∴ Potential difference increases.
Question 4
4 / -1

A parallel plate capacitor is charged and then isolated. What is the effect of increasing the plate separation on charge, potential, capacitance, respectively?

A
constant, decreases, decreases

B
increases, decreases, decreases

C
constant, decreases, increases

D
constant, increases, decreases

Solution

As the capacitor is isolated after charging, charge on it remains constant. Plate separation increases d, decreases C = ϵ₀A/d and hence increases potential V = q/C.
Question 5
4 / -1

A parallel plate capacitor is connected to a battery as shown in figure. Consider two situations

(i) key KK is kept closed and plates of capacitors are moved apart using insulating handle
(ii) key KK is opened and plates of capacitors are moved apart using insulating handle
Which of the following statements is correct?

A
In (i), Q remains same but C changes

B
In (ii) V remains same but C changes

C
In (i) V remains same and hence Q changes

D
In (ii) both Q and V changes

Solution

When key K is kept closed, condenser C is charged to potential V. When plates of capacitors are moved apart, its capacitance, C = ϵ_oA/d decreases.
As potential of condenser remains same, charge Q = CV decreases. So option is correct. Once key K is closed, condenser gets charged, Q = CV.
Now, if key K is opened, battery is disconnected, no more charging can occur i.e. Q remains same.
As plates or capacitor are moved apart, its capacity C = ϵ_oA/d decreases.
Therefore, its potential, V = q/C increases