Q#1
The capacitor of capacitance C is charged to a potential V. The flux of the electric field through a closed surface enclosing the capacitor is
(a) $\frac{CV}{\epsilon_0}$
(b) $\frac{2CV}{\epsilon_0}$
(c) $\frac{CV}{2\epsilon_0}$
(d) Zero
Answer: (d)
The charge on the plates of a capacitor are +Q and -Q. Hence the net charge (q) on a capacitor is zero.
The flux of electric field through a closed surface enclosing the capacitor = $\frac{q}{\epsilon_0}$ = $\frac{0}{\epsilon_0}$ = 0.
Hence the option (d) is correct.
Q#2
Two capacitors each having capacitance C and breakdown voltage V are joined in series. The capacitance and the breakdown voltage of the combination will be
(a) 2C and 2V
(b) $\frac{C}{2}$ and $\frac{V}{2}$2
(c) 2C and $\frac{V}{2}$
(d) $\frac{C}{2}$ and 2V.
Answer: (d)
Since the capacitors are joined in series, the equivalent capacitance, C' is given as,
$\frac{1}{C'}=\frac{1}{C}+\frac{1}{C}=\frac{2}{C}$
C' = $\frac{C}{2}$.
But the voltage in the series combination is added up. Hence the breakdown voltage of the series combination = V+V = 2V. Hence the option (d) is correct.
Q#3
If the capacitors in the previous question are joined in parallel, the capacitance and the breakdown voltage of the combination will be
(a) 2C and 2V
(b) C and 2V
(c) 2C and V
(d) C and V
Answer: (c)
Capacitance of a parallel combination = C+C = 2C.
Since the plates of both capacitors are connected to the same positive and negative terminals of a battery, the potential of the positive and negative terminals of each capacitor is the same. This results in the same potential difference across the plates. Hence the breakdown voltage of the combination is V.
So the option (c) is correct.
Q#4
The equivalent capacitance of the combination shown in figure (31-Q1) is
(a) C
(b) 2C
(c) $\frac{C}{2}$
(d) none of these.
Answer: (b)
Let us name the points given in the circuits.
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| The Figure for Q#4 |
the potential difference between these two points is zero. The capacitor in this line will have no effect on the circuit.
Rest two capacitors are connected in parallel. Hence the equivalent capacitance = 2C.
So the option (b) is true.
Q#5
A dielectric slab is inserted between the plates of an isolated capacitor. The force between the plates will
(a) increase
(b) decrease
(c) remain unchanged
(d) become zero.
Answer:
The induced charges on the surface of the inserted dielectric induce an extra electric field inside the dielectric but it has no effect on the outside. So the electric field by the positive plate near the negative plate remains unaffected. Hence the force between the plates will remain unchanged.
Option (c) is correct.
Q#6
The energy density in the electric field created by a point charge falls off with the distance from the point charge as
(a) 1/r
(b) 1/r²
(c) 1/r³
(d) 1/r⁴
Answer: (d)
The energy density of an electric field = $\frac{1}{2}\epsilon_0 E^2$, where E is the intensity of the electric field.
The electric field at a distance r from a point charge q,
E = $\frac{q}{4\pi \epsilon_0 r^2}$
Putting this value of E above we get, energy density
= $\frac{1}{2}\epsilon_0 \left(\frac{q}{4\pi \epsilon_0 r^2}\right)^2$
= $\left(\frac{q^2}{8\pi \epsilon_0^2}\right) \times \frac{1}{r^4}$
= (A constant) $\times \frac{1}{r^4}$
Hence the energy density falls off with the distance from point charge as $\frac{1}{r^4}$.
Option (d) is correct.
Q#7
A parallel plate capacitor has plates of unequal area. The larger plate is connected to the positive terminal of the battery and the smaller plate to its negative terminal. Let Q₊ and Q₋ be the charges appearing on the positive and negative plates respectively.
(a) Q₊ > Q₋
(b) Q₊ = Q₋
(c) Q₊ < Q₋
(d) The information is not sufficient to decide the relation between Q₊ and Q₋.
Answer: (b)
Total charge inside a battery remains constant. When the terminals of the battery are connected to the plates of a capacitor, a positive charge is sent to the plate connected to the positive terminal. An equal amount of negative charge is sent to the plate connected to the negative terminal of the battery. This makes the total charge inside the battery unchanged. So, the charges appearing on the plates have no relation to their areas. Both the plates have the same amount of charges but opposite in nature.
Option (b) is correct.
Q#8
A thin metal plate P is inserted between the plates of a parallel plate capacitor of capacitance C in such a way that its edges touch the two plates figure(31-Q2). The capacitance now becomes
(a) $\frac{C}{2}$
(b) 2C
(c) 0
(d) ∞.
Answer: (d)
The capacitance of a parallel plate capacitor,
$C=\frac{\epsilon_0 A}{d}$, where d is the distance between the plates.
In this case, the metal plate P touches both the plates, hence the distance between the plates d = 0. This makes capacitance C = ∞.
Hence the option (d) is correct.
Q#9
Figure (31-Q3) shows two capacitors connected in series and joined to a battery. The graph shows the variation in potential as one moves from left to right on the branch containing the capacitors.
(a) C₁ > C₂
(b) C₁ = C₂
(c) C₁ < C₂
(d) The information is not sufficint to decide the relation between C₁ and C₂.
Answer: (c)
From the graph, it is clear that the potential difference across the capacitor C₁ is greater than the potential difference across C₂. The charge on each capacitor will be the same.
Capacitance C = $\frac{Q}{V}$, so the capacitor with a greater potential difference will have lower capacitance.
So, C₁ < C₂. Option (c) is correct.
Q#10
Two metal plates having charges Q, -Q face each other at some separation and are dipped into an oil tank. If the oil is pumped out, the electric field between the plates will
(a) increase
(b) decrease
(c) remain the same
(d) become zero.
Answer: (a)
A plate of a capacitor is a charged conducting surface.
The electric field near a charged conducting surface is
E = $\frac{\sigma}{\epsilon_0}$
Since the charge is the same on the plates in oil and air, 𝜎 is unchanged.
It is the permittivity εₒ of air which is less than oil. Hence the electric field between the plates in oil will be lesser than air.
So when the oil is pumped out, the electric field between the plates will increase.
Option (a) is correct.
Q#11
Two metal spheres of capacitances 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) $\frac{Q_1}{Q_2}< \frac{C_1}{C_2}$
(b) $\frac{Q_1}{Q_2}= \frac{C_1}{C_2}$
(c) $\frac{Q_1}{Q_2}> \frac{C_1}{C_2}$
(d) $\frac{Q_1}{Q_2}= \frac{C_2}{C_1}$
Answer: (b)
When the spheres are put in contact, both acquire the same potential, say V.
Now Q₁ = C₁V, and Q₂ =C₂V.
Hence $\frac{Q_1}{Q_2}=\frac{C_1}{C_2}$. Option (b) is correct.
Q#12
Three capacitors of capacitances 6 µF each are available. The minimum and maximum capacitances, which may be obtained are
(a) 6 µF and 18 µF
(b) 3 µF and 12 µF
(c) 2 µF and 12 µF
(d) 2 µF and 18 µF.
Answer: (d)
Minimum capacitance can be obtained by joining them in series.
$\frac{1}{C}=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{3}{6}=\frac{1}{2}$
C = 2 µF.
Maximum capacitance can be obtained by joining them in parallel. Capacitance in this case,
C = 6 +6 +6 =18 µF.
Option (d) is correct.
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