Tag: physics

Questions Related to physics

Find the potential at a point due to a positive charge of $100\mu C$ at a distance of $10\ m$ in a medium of dielectric constant $9$.

parallel plate capacitor electrostatic potential and capacitance electrostatics physics

  1. $10^{7}V$.

  2. $10^{4}V$.

  3. $10^{5}V$.

  4. $10^{6}V$.

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Correct Option: B

A capacitor of capacitance 2 $\mu $ F is charged to a potential difference of 12 V. It is then connected across inductor of inductance 0.6 mH. The current in the circuit when the potential difference across the capacitor 6 V is :

parallel plate capacitor electrostatic potential and capacitance electrostatics physics

  1. 3.6 A

  2. 2.4 A

  3. 1.2 A

  4. 0.06 A

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Correct Option: D

The parallel plates of capacitor are charged to a potential difference of 320 volts and are then connected across a resistor. The potential difference across the capacitor decays exponentially with time. Alter 1 second the potential difference between the plates of the capacitor is 240 volts then after 2 seconds the potential difference between the plates will be -

parallel plate capacitor electrostatic potential and capacitance electrostatics physics

  1. 200 V

  2. 180 V

  3. 160 V

  4. 140 V

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Correct Option: B

When charge is supplied in a system where the capacitor are connnected in parallel. the charge will be distributed

parallel plate capacitor electrostatic potential and capacitance electrostatics physics

  1. Equally to all the capacitor

  2. indirect ratio to the capacitance value

  3. in an inverse ratio to the capacitance values

  4. both (b) and (c)

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Correct Option: C

Two identical capacitors are connected in parallel across a potenial difference V. after they are fully charged, the positive plate of first capacitor is connected to negative plate of second and negative plate of first is connected to positive plate of other. The loss of energy will be

parallel plate capacitor electrostatic potential and capacitance electrostatics physics

  1. $\dfrac { 1 }{ 2 } { CV }^{ 2 }$

  2. ${ CV }^{ 2 }$

  3. $\dfrac { 1 }{ 4 } { CV }^{ 2 }$

  4. Zero

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Correct Option: C

A simple pendulum of mass m charged negatively to q coulomb oscillates with a time period T in a downward electric field E such that mg > qE. If the electric field is withdrawn, the new time period :

parallel plate capacitor electrostatic potential and capacitance electrostatics physics

  1. $=$T

  2. $>$T

  3. $<$T

  4. any of the above three is possible

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Correct Option: C
Explanation:

Time period in the absence of electric field $T' = 2 \pi \displaystyle \sqrt{\frac{l}{g}}$

In the presence of electric field
$g _{eff}=(mg - qE) / m$
Therefore $T = 2 \pi \displaystyle \sqrt{\frac{l}{(mg - qE) / m}}$
$\displaystyle T'  = 2\pi \sqrt{\frac{l}{g}}$
or $T' < T$

Among two discs $A$ and $B$, first have radius $10\ cm$ and charge ${10}^{-6}\ \mu C$ and second have radius $30\ cm$ and charge ${10}^{-5}C$. When they are touched, charge on both ${q} _{A}$ and ${q} _{B}$ respectively will be :

parallel plate capacitor electrostatic potential and capacitance electrostatics physics

  1. ${q} _{A}=2.75\mu C,{q} _{B}=3.15\mu C$

  2. ${q} _{A}=1.09\mu C,{q} _{B}=1.53\mu C$

  3. ${q} _{A}={q} _{B}=5.5\mu C$

  4. none of these

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Correct Option: C

Two metal pieces having a potential difference of 800 V are 0.02 m apart horizontally. A particle of mass $1.96\times 10^{-15}kg$ is suspended in equilibrium between the plates. If e is the elementary charge, then charge on the particle is

parallel plate capacitor electrostatic potential and capacitance electrostatics physics

  1. 8

  2. 6

  3. 0.1

  4. 3

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Correct Option: D
Explanation:

Force due to gravity on the particle is $F _g=mg$
Force due to field E between the metal plates is $\displaystyle F _e=qE=(ne)\frac{V}{d}$.
In equilibrium, $\displaystyle F _g=F _e \Rightarrow mg=(ne)\frac{V}{d}$


$\displaystyle \therefore n=\dfrac{mgd}{eV}$

$=\dfrac{1.96\times 10^{-15}\times 9.8\times 0.02}{1.6\times 10^{-19}\times 800}=3$

Consider two bodies A and B of same capacitance.  If charge of -10C flows from body A to body B, then

parallel plate capacitor electrostatic potential and capacitance electrostatics physics

  1. the potential of body A increases.

  2. the potential of body B decreases

  3. the magnitude of change in potential in both bodies is same.

  4. All the above

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Correct Option: D
Explanation:

As we know $ V = \dfrac{Q}{C}$ , where letters have their respective meanings.

So, if charge of $-10C$  (Negative Charge) flows from body A to body B, then the potential of body A increases and the potential of body B decreases. Also the magnitude of change in potential in both bodies is same as A and B are of same capacitance.
Therefore, D is correct option.

Which of the following is true about field between parallel charged plates?

parallel plate capacitor electrostatic potential and capacitance electrostatics physics

  1. It is strongest between the plates

  2. It is strongest near the positive plate

  3. It is strongest near the negative plate

  4. The field is constant between the plates

  5. The field is variable, therefore the strong point also varies

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Correct Option: D
Explanation:

The field is constant between the plates , because we dealing with a system of two charged plates.we know the electric field between two charged plates (suppose both have positive charge) is given by 

                   $E=\frac{1}{2\varepsilon _{0}}\left(\sigma _{1}-\sigma _{2}\right)$         where $\sigma=$ surface charge density
we can see that electric field doesn't depend upon distance from plates , therefore it is constant.