In the following circuit, the equivalent capacitance between terminal $A$and terminal $B$ is :
- A$1 \mu \mathrm{F}$
- B$0.5 \mu \mathrm{F}$
- C$4 \mu \mathrm{F}$
- ✓$2 \mu \mathrm{F}$
Answer: D.
View full solution →Question types
2. Electric potential and capacitance question types
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Sample Questions
2. Electric potential and capacitance questions
One sample from each question group in this chapter. Select any group above to see the full set with answer keys.
If the plates of a parallel plate capacitor connected to a battery are moved close to each other, then
$A$. the charge stored in it, increases.
$B$. the energy stored in it, decreases.
$C$. its capacitance increases.
$D$. the ratio of charge to its potential remains the same.
$E$. the product of charge and voltage increases.
Choose the most appropriate answer from the options given below:
- ✓$A, C$ and $E$ only
- B$B, D$ and $E$ only
- C$A, B$ and $C$ only
- D$A, B$ and $E$ only
Answer: A.
View full solution →Given below are two statements: one is labelled as Assertion $A$ and the other is labelled as Reason $R$.
Assertion $A$: The potential ( $V$ ) at any axial point, at $2 \mathrm{~m}$ distance ( $r$ ) from the centre of the dipole of dipole moment vector $\vec{P}$ of magnitude, $4 \times 10^{-6} \mathrm{C} \mathrm{m}$, is $\pm 9 \times 10^3 \mathrm{~V}$.
(Take $\frac{1}{4 \pi \in_0}=9 \times 10^9 \mathrm{Sl}$ units)
Reason $R$: $V= \pm \frac{2 P}{4 \pi \in_0 r^2}$, where $r$ is the distance of any axial point, situated at $2 \mathrm{~m}$ from the centre of the dipole.
In the light of the above statements, choose the correct answer from the options given below:
- ABoth $A$ and $R$ are true and $R$ is $NOT$ the correct explanation of $A$.
- ✓$A$ is true but $R$ is false.
- C$A$ is false but $R$ is true.
- DBoth $A$ and $R$ are true and $R$ is the correct explanation of $A$.
Answer: B.
View full solution →A thin spherical shell is charged by some source. The potential difference between the two points $C$ and $P$ (in $V$) shown in the figure is:
(Take $\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9$ $SI$ units)
- A$1 \times 10^5$
- B$0.5 \times 10^5$
- ✓Zero
- D$3 \times 10^5$
Answer: C.
View full solution →An electric dipole is placed as shown in the figure.The electric potential (in $10^2\,V$ ) at point $P$ due to the dipole is $\left(\in_0=\right.$ permittivity of free space and $\left.\frac{1}{4 \pi \in_0}=K\right)$ :
- A$\left(\frac{8}{3}\right) qK$
- ✓$\left(\frac{3}{8}\right) qK$
- C$\left(\frac{5}{8}\right) qK$
- D$\left(\frac{8}{5}\right) qK$
Answer: B.
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