Question 2011 Mark
The electric potential at a point on the equatorial line of an electric dipole is:
Answer
View full question & answer→- None of these.
Questions · Page 5 of 6
M.C.Q (1 Marks)
Question 2021 Mark
Which of the following is / are true about the principle of Van de Graaff generator?
Answer
View full question & answer→- Both (a) and (b).
Question 2031 Mark
In a charged capacitor, the energy resides:
Answer
View full question & answer→- In the field between the plates.
Question 2041 Mark
The magnitude of the electric field is given by the change in magnitude of potential per unit displacement _____ to the equipotential surface at the point.
Answer
View full question & answer→- Both a and b.
Question 2051 Mark
When was Van de Graff generator invented and by whom?
Answer
Van de Graff generator was invented by Robert Jemison Van de Graff on November 28, 1933. Robert Jemison invented the Van de Graff generator, which is a kind of high-voltage electrostatic generator that accelerates particles, while he was doing his PhD in Princeton University.
View full question & answer→- 1933, Robert Van de Graff
Van de Graff generator was invented by Robert Jemison Van de Graff on November 28, 1933. Robert Jemison invented the Van de Graff generator, which is a kind of high-voltage electrostatic generator that accelerates particles, while he was doing his PhD in Princeton University.
Question 2061 Mark
The dipole moment per unit volume is called as:
Answer
View full question & answer→- Polarisation.
Question 2071 Mark
A parallel plate air capacitor has capacity ′C′ farad, potential ′V′ volt and energy ′E′ joule. When the gap between the plates is completely filled with dielectric.
Answer
A parallel- plate capacitor with a dielectric. The electric field is reduced between the plates because the dielectric material is polarized, producing an opposing field. When there is a dielectric, the potential is also reduced because potential is inversely proportional to dielectric
View full question & answer→- Both V and E decrease
A parallel- plate capacitor with a dielectric. The electric field is reduced between the plates because the dielectric material is polarized, producing an opposing field. When there is a dielectric, the potential is also reduced because potential is inversely proportional to dielectric
Question 2081 Mark
1 Volt = _____________ style-type: lower- alpha;”>
Answer
View full question & answer→- 1 Joule/ 1 Coulomb
Question 2091 Mark
From a point charge, there is a fixed point A. At A, there is an electric field of 500V/ m and potential difference of 3000V. Distance between point charge and A will be:
Answer
View full question & answer→- 6m.
Question 2101 Mark
1 Pico farad =
Answer
View full question & answer→- 10-12.
Question 2111 Mark
A capacitor is a system of two conductors separated by _____.
Answer
View full question & answer→- An insulators.
Question 2121 Mark
Capacity of a parallel plate condenser can be increased by:
Answer
View full question & answer→- Decreasing the distance between the plates.
Question 2131 Mark
Electrostatic potential is ______ throughout the volume of the conductor and has _____ value on its surface.
Answer
View full question & answer→- Constant, same.
Question 2141 Mark
What is the electric field in the cavity of a hollow charged conductor?
Answer
By Gauss’s theorem, the charge enclosed by the gaussian surface is zero. Consequently, the electric field must be zero at every point inside the cavity. Then, the entire excess charge lies on its surface.
View full question & answer→- Zero
By Gauss’s theorem, the charge enclosed by the gaussian surface is zero. Consequently, the electric field must be zero at every point inside the cavity. Then, the entire excess charge lies on its surface.
Question 2151 Mark
$1$ electron volt $= ........J.$
Answer
View full question & answer→$1$ electron volt is the amount of work done if an electron is passed through a potential difference of $1V.$ Therefore the work done $= 1V \times$ charge of an electron $= 1.602 \times 10^{-19} J.$ But it is a small quantity and hence we use kilo electron volt and mega electron volt in practical.
Question 2161 Mark
Following operations can be performed on a capacitor:
$X -$ connect the capacitor to a battery of emf $\epsilon.$
$Y -$ disconnect the battery.
$Z -$ reconnect the battery with polarity reversed.
$W -$ insert a dielectric slab in the capacitor.
$X -$ connect the capacitor to a battery of emf $\epsilon.$
$Y -$ disconnect the battery.
$Z -$ reconnect the battery with polarity reversed.
$W -$ insert a dielectric slab in the capacitor.
Answer
If the potential is held constant, that is, the battery remains attached to the circuit, then the charge on the capacitor increases by a factor of $K$ on inserting a dielectric of a dielectric constant $K$ between the plates of the capacitor.
Mathematically,
$q = Kq_{0 }$
Here, $q_0$ and $q$ are the charges without dielectric and with dielectric, respectively.
The amount of charge stored does not depend upon the polarity of the plates.
Thus, the charge appearing on the capacitor is greater after the action $\text{XWY}$ than after the action $\text{XYZ}.$
Justification of option $(c):$
Since the battery is disconnected before inserting a dielectric, the amount of charge remains constant, that is, $q= q_{0},$ because after the battery is disconnected, the capacitor gets no source to store charge from. In other words, the capacitor is now an isolated system where the amount of charge is conserved and so is the energy U as $\frac{1}{2}\text{q}\epsilon.$
Hence, inserting a dielectric after disconnecting the battery will not bring any change in the amount of charge stored in the capacitor.
So, the energy stored in the capacitor will also not change after the action $\text{XYW}.$
However, during the action $\text{WXY}$, the amount of charge that will get stored in the capacitor will get increased by a factor of $K$, as the battery is disconnected after inserting a dielectric between the plates of the capacitor and the energy stored will also get multiplied by a factor of $K.$
Thus, the electric energy stored in the capacitor is greater after the action $\text{WXY}$ than after the action $\text{XYW}.$
Justification of option $(d):$
The electric field between the plates $E$ depends on the potential across the capacitor and the distanced between the plates of the capacitor.
Mathematically,
$\text{E}=\frac{\epsilon}{\text{d}}$
In either case, that is, during actions $XW$ and $WX,$ the potential remains the same, that is, $\epsilon.$ Thus, the electric field $E$ remains the same.
Denial of option $(a):$
During the action $\text{XYZ},$ the battery has to do extra work equivalent to $\frac{1}{2}\text{CV}^2$ to change the polarity of the plates of the capacitor. In other words, the total work to be done by the battery will be $\frac{1}{2}\text{CV}^2+\frac{1}{2}\text{CV}^2.$
This extra work done will be dissipated as heat energy.
Thus, thermal energy is developed. However, the stored electric energy remains unchanged, that is, $\frac{1}{2}\text{CV}^2.$
View full question & answer→- The charge appearing on the capacitor is greater after the action $\text{XWY}$ than after the action $\text{XWY}$
- The electric energy stored in the capacitor is greater after the action $\text{WXY}$ than after the action $\text{XYW}$.
- The electric field in the capacitor after the action $XW$ is the same as that after $WX.$
If the potential is held constant, that is, the battery remains attached to the circuit, then the charge on the capacitor increases by a factor of $K$ on inserting a dielectric of a dielectric constant $K$ between the plates of the capacitor.
Mathematically,
$q = Kq_{0 }$
Here, $q_0$ and $q$ are the charges without dielectric and with dielectric, respectively.
The amount of charge stored does not depend upon the polarity of the plates.
Thus, the charge appearing on the capacitor is greater after the action $\text{XWY}$ than after the action $\text{XYZ}.$
Justification of option $(c):$
Since the battery is disconnected before inserting a dielectric, the amount of charge remains constant, that is, $q= q_{0},$ because after the battery is disconnected, the capacitor gets no source to store charge from. In other words, the capacitor is now an isolated system where the amount of charge is conserved and so is the energy U as $\frac{1}{2}\text{q}\epsilon.$
Hence, inserting a dielectric after disconnecting the battery will not bring any change in the amount of charge stored in the capacitor.
So, the energy stored in the capacitor will also not change after the action $\text{XYW}.$
However, during the action $\text{WXY}$, the amount of charge that will get stored in the capacitor will get increased by a factor of $K$, as the battery is disconnected after inserting a dielectric between the plates of the capacitor and the energy stored will also get multiplied by a factor of $K.$
Thus, the electric energy stored in the capacitor is greater after the action $\text{WXY}$ than after the action $\text{XYW}.$
Justification of option $(d):$
The electric field between the plates $E$ depends on the potential across the capacitor and the distanced between the plates of the capacitor.
Mathematically,
$\text{E}=\frac{\epsilon}{\text{d}}$
In either case, that is, during actions $XW$ and $WX,$ the potential remains the same, that is, $\epsilon.$ Thus, the electric field $E$ remains the same.
Denial of option $(a):$
During the action $\text{XYZ},$ the battery has to do extra work equivalent to $\frac{1}{2}\text{CV}^2$ to change the polarity of the plates of the capacitor. In other words, the total work to be done by the battery will be $\frac{1}{2}\text{CV}^2+\frac{1}{2}\text{CV}^2.$
This extra work done will be dissipated as heat energy.
Thus, thermal energy is developed. However, the stored electric energy remains unchanged, that is, $\frac{1}{2}\text{CV}^2.$
Question 2171 Mark
The factor by which the capacitance increases from its vacuum value when the dielectric is inserted fully between the plates of a capacitor, is called as:
Answer
View full question & answer→- Dielectric constant of the substance.
Question 2181 Mark
The capacitance is measured in:
Answer
View full question & answer→- Farad.
Question 2191 Mark
In parallel combination of capacitors, the effective capacitance:
Answer
View full question & answer→- Increases.
Question 2201 Mark
A point charge q is rotated along a circle in the electric field generated by another point charge Q. The work done by the electric field on the rotating charge in one complete revolution is_______?
Answer
The net displacement round one complete circle is 0.
View full question & answer→- zero
The net displacement round one complete circle is 0.
Question 2211 Mark
In case of a Van de Graaff generator, the breakdown field of air is:
Answer
View full question & answer→SELF
Question 2221 Mark
The work done by the external force in bringing the charge q from infinity to a point is called as:
Answer
View full question & answer→- Potential energy due to charge q at that point.
Question 2231 Mark
The dimensions of capacitance are:
Answer
View full question & answer→- [M-1 L-2 T-4 A2]
Question 2241 Mark
The forces in which sum of kinetic energy and potential energy is conserved are called as:
Answer
View full question & answer→- Conservative forces.
Question 2251 Mark
An electric dipole is placed in a uniform electric field. The net electric force on the dipole.
Answer
Net Electric force $= F_A + F_B$
$= -qE + qE$
$= 0$
View full question & answer→Net Electric force $= F_A + F_B$
$= -qE + qE$
$= 0$
Question 2261 Mark
What will be the nature of equipotential surfaces due to a point charge, situated at infinity?
Answer
If a point charge is situated at infinity, the electric field lines coming out of it will be in the form of parallel straight lines. As we know that field lines cut the equipotential surfaces orthogonally, therefore the equipotential surfaces must be plane surfaces. They can be considered the surface of a sphere of infinite radius.
View full question & answer→- Plane surface
If a point charge is situated at infinity, the electric field lines coming out of it will be in the form of parallel straight lines. As we know that field lines cut the equipotential surfaces orthogonally, therefore the equipotential surfaces must be plane surfaces. They can be considered the surface of a sphere of infinite radius.
Question 2271 Mark
A capacitor becomes a perfect insulator for:
Answer
A capacitor becomes a perfect insulator for direct current as the regular supply of current charges capacitor and the it behaves as open circuit, where as in A.C. the current being variable in sign and magnitude does not charge capacitor ever it goes through a repetitive process of charging and discharging and hence it never behaves as open circuit.
View full question & answer→- Direct current
A capacitor becomes a perfect insulator for direct current as the regular supply of current charges capacitor and the it behaves as open circuit, where as in A.C. the current being variable in sign and magnitude does not charge capacitor ever it goes through a repetitive process of charging and discharging and hence it never behaves as open circuit.
Question 2281 Mark
If in a parallel plate capacitor, which is connected to a battery, we fill dielectrics in whole space of its plates, then which of the following increases?
Answer
Since battery remains connected so P.D. between the plates is constant. But as we introduce the dielectric the capacitance increases and hence charge increases.
View full question & answer→- Q and C
Since battery remains connected so P.D. between the plates is constant. But as we introduce the dielectric the capacitance increases and hence charge increases.
Question 2291 Mark
What is the induced dipole moment developed per unit volume of a dielectric when placed in an external electric field called?
Answer
The induced dipole moment developed per unit volume of a dielectric when placed in an external electric field is called polarization density. It may be defined as the charge induced per unit surface area.
View full question & answer→- Polarisation density.
The induced dipole moment developed per unit volume of a dielectric when placed in an external electric field is called polarization density. It may be defined as the charge induced per unit surface area.
Question 2301 Mark
Dielectric constant is the:
Answer
View full question & answer→- Dimensions less quantity.
Question 2311 Mark
The amount of work done in moving a unit positive charge from infinity to a given point is known as:
Answer
Electric potential may be defined as the amount of work done in moving a unit positive charge from infinity to a given point.
$\text{V}=\frac{\text{W}}{\text{q}}$
View full question & answer→- Electric potential
Electric potential may be defined as the amount of work done in moving a unit positive charge from infinity to a given point.
$\text{V}=\frac{\text{W}}{\text{q}}$
Question 2321 Mark
Dielectrics are ____ substances.
Answer
View full question & answer→- Non conducting.
Question 2331 Mark
It becomes possible to define potential at a point in an electric field because electric field:
Answer
View full question & answer→- Is a conservative field.
Question 2341 Mark
Three capacitors each of capacity $ 4μ\text{F}$ are to be connected in such a way that the effective capacitance is $6μ\text{F}$. This can be done by:
Answer
View full question & answer→- Connecting two in series and one in parallel.
Question 2351 Mark
Three capacitors, $ 3μ\text{F}, 6μ\text{F}$ and 6μF are connected in series to a source of 120V. The potential difference, in volts, across the $3μ\text{F}$ capacitor will be:
Answer
The equivalent capacitance of the two $6μ\text{F} $ and $6μ\text{F} $ capacitors in series is $3μ\text{F} $.
Hence the potential across the two capacitors, original $3μ\text{F} $ capacitor and the equivalent $3μ\text{F} $ capacitor is divided equally.
View full question & answer→- 60
The equivalent capacitance of the two $6μ\text{F} $ and $6μ\text{F} $ capacitors in series is $3μ\text{F} $.
Hence the potential across the two capacitors, original $3μ\text{F} $ capacitor and the equivalent $3μ\text{F} $ capacitor is divided equally.
Question 2361 Mark
How many $6μ\text{F},$ 200V condensers are needed to make a condenser of $18μ\text{F},$ 600V?
Answer
Place three 200V, $6μ\text{F},$ capacitors in series to get 1 equivalent 600V, $2μ\text{F},$ capacitor. Now place 9 of these equivalent 600V, $2μ\text{F},$capacitors in parallel to obtain an equivalence of 18μF at 600 Volts. All this requires a total of 27 $6μ\text{F},$ capacitors. Nine rows connected in parallel with 3 capacitors connected in series in each row.
View full question & answer→- 27
Place three 200V, $6μ\text{F},$ capacitors in series to get 1 equivalent 600V, $2μ\text{F},$ capacitor. Now place 9 of these equivalent 600V, $2μ\text{F},$capacitors in parallel to obtain an equivalence of 18μF at 600 Volts. All this requires a total of 27 $6μ\text{F},$ capacitors. Nine rows connected in parallel with 3 capacitors connected in series in each row.
Question 2371 Mark
Consider a uniform electric field in the $\hat{\text{Z}}$ direction. The potential is a constant:
Answer
$\text{E}=-\frac{\text{dV}}{\text{dr}}$
Electric potential decreases inf the direction of electric field. The direction of electric field is always perpendicular to one equipotential surface maintained at high electrostatic potential to other equipotential surface maintained at low electrostatic potential.
The electric field in z-direction suggest that equipotential surfaces are in x-y plane. Therefore the potential is a constant for any x for a given z, for any y for a given z and on the x-y plane for a given z.
View full question & answer→- For any x for a given z.
- For any y for a given z.
- On the x-y plane for a given z.
$\text{E}=-\frac{\text{dV}}{\text{dr}}$
Electric potential decreases inf the direction of electric field. The direction of electric field is always perpendicular to one equipotential surface maintained at high electrostatic potential to other equipotential surface maintained at low electrostatic potential.
The electric field in z-direction suggest that equipotential surfaces are in x-y plane. Therefore the potential is a constant for any x for a given z, for any y for a given z and on the x-y plane for a given z.
Question 2381 Mark
The minimum number of condensers each of capacitance of $2μ\text{F},$ in order to obtain resultant capacitance of $5μ\text{F},$ will be:
Answer
We can obtain an equivalence capacitance of $5μ\text{F},$ by connecting minimum 4 capacitance of each $2μ\text{F},$only in the way as shown in the figure.
View full question & answer→- 4
We can obtain an equivalence capacitance of $5μ\text{F},$ by connecting minimum 4 capacitance of each $2μ\text{F},$only in the way as shown in the figure.
Question 2391 Mark
In a parallel plate capacitor, the capacity increases if:
Answer
Hint:- Check the dependence of capacitance on certain quantities.
In a parallel plate capacitor, the capacitance is $\text{C}=\frac{\text{k}\epsilon_0\text{A}}{\text{d}};$
Where, k is the dielectric constant, $\epsilon_0$ is the permittivity constant, A is the area of the conduction and d is the distance between plates.
From here we can see C is directly proportional to k, $\epsilon_0$,A and inversely proportional to d.
View full question & answer→- Area of the plate is increased.
Hint:- Check the dependence of capacitance on certain quantities.
In a parallel plate capacitor, the capacitance is $\text{C}=\frac{\text{k}\epsilon_0\text{A}}{\text{d}};$
Where, k is the dielectric constant, $\epsilon_0$ is the permittivity constant, A is the area of the conduction and d is the distance between plates.
From here we can see C is directly proportional to k, $\epsilon_0$,A and inversely proportional to d.
Question 2401 Mark
Figure shows some equipotential lines distributed in space. A charged object is moved from point A to point B.
Answer
The positively charged particle experiences electrostatic force along the direction of electric field, hence moves in the direction opposite to electric field. Thus, the work done by the electric field on the charge will be negative. We know
$\text{W}_\text{electrical}=-\Delta\text{U}=-\text{q}\Delta\text{V}=\text{q}(\text{V}_\text{Intial}-\text{V}_\text{final})$
Here initial and final potentials are same in all three cases and same charge is moved, so work done is same in all three cases.
View full question & answer→- The work done is the same in Fig. (i) Fig. (ii) and Fig. (iii).
- The density of the equipotential lines gives an idea about the magnitude of electric field. Higher the density, larger the field strength.
- The direction of electric field is perpendicular to the equipotential surfaces or lines.
- The equipotential surfaces produced by a point charge or a spherically charge distribution are a family of concentric spheres.
- For a uniform electric field, the equipotential surfaces are a family of plane perpendicular to the field lines.
- A metallic surface ofany shape is an equipotential surface.
- Equipotential surfaces can never cross each other.
- The work done in moving a charge along an equipotential surface is always zero.
The positively charged particle experiences electrostatic force along the direction of electric field, hence moves in the direction opposite to electric field. Thus, the work done by the electric field on the charge will be negative. We know
$\text{W}_\text{electrical}=-\Delta\text{U}=-\text{q}\Delta\text{V}=\text{q}(\text{V}_\text{Intial}-\text{V}_\text{final})$
Here initial and final potentials are same in all three cases and same charge is moved, so work done is same in all three cases.
Question 2411 Mark
If the capacitors in the previous question are joined in parallel, the capacitance and the breakdown voltage of the combination will be:
Answer
View full question & answer→In a parallel combination of capacitors, the potential difference across the capacitors remain the same, as the right$-$hand$-$side plates and the left$-$hand$-$side plates of both the capacitors are connected to the same terminals of the battery. Therefore, the potential remains the same, that is, $V.$
For the parallel combination of capacitors, the capacitance is given by
$C_{eq} = C_1 + C_2$
Here,
$C_1 = C_2 = C$
$\therefore C_{eq} = 2C$
For the parallel combination of capacitors, the capacitance is given by
$C_{eq} = C_1 + C_2$
Here,
$C_1 = C_2 = C$
$\therefore C_{eq} = 2C$
Question 2421 Mark
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:
Answer
View full question & answer→The charge induced on the plates of a capacitor is independent of the area of the plates.
$\therefore Q_{+ }= Q_-$
$\therefore Q_{+ }= Q_-$
Question 2431 Mark
(1): The dielectric medium between the plates of a parallel plate capacitor lowers the potential difference between the plates without a battery.
(2): The maximum electric field that a dielectric can withstand without causing it to break down is dielectric strength.
(2): The maximum electric field that a dielectric can withstand without causing it to break down is dielectric strength.
Answer
Consider a capacitor with charge density $σ.$
The potential between its two plates is given by $\frac{\sigma\text{d}}{\epsilon_0}$
When a dielectric is inserted, the electric field inside the capacitor decrease decreasing the potential between the two plates of capacitor.
However, this is nothing to the dielectric strength of the dielectric.
View full question & answer→- Both 1 and 2 are true, 2 is not correct explanation of 1
Consider a capacitor with charge density $σ.$
The potential between its two plates is given by $\frac{\sigma\text{d}}{\epsilon_0}$
When a dielectric is inserted, the electric field inside the capacitor decrease decreasing the potential between the two plates of capacitor.
However, this is nothing to the dielectric strength of the dielectric.
Question 2441 Mark
Figure 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:
Answer
Region $AB$ shows the potential difference across capacitor $C_{1}$ and region $CD$ shows the potential difference across capacitor $C_2.$ Now, we can see from the graph that region $AB$ is greater than region $CD.$ Therefore, the potential difference across capacitor $C_1$ is greater than that across capacitor $C_2.$
$\because$ Capacitance, $\text{C}=\frac{\text{Q}}{\text{V}}$
$\therefore\text{C}_1<\text{C}_2(Q$ remains the same in series connection$).$
View full question & answer→Region $AB$ shows the potential difference across capacitor $C_{1}$ and region $CD$ shows the potential difference across capacitor $C_2.$ Now, we can see from the graph that region $AB$ is greater than region $CD.$ Therefore, the potential difference across capacitor $C_1$ is greater than that across capacitor $C_2.$
$\because$ Capacitance, $\text{C}=\frac{\text{Q}}{\text{V}}$
$\therefore\text{C}_1<\text{C}_2(Q$ remains the same in series connection$).$
Question 2451 Mark
One Volt is equal to:
Answer
The volt is a measure of electric potential. One volt is defined as the difference in electric potential between two points of a conducting wire when an electric current of one ampere dissipates one watt of power between those points. It is also equal to the potential difference between two parallel, infinite planes spaced 1 meter apart that create an electric field of 1 newton per coulomb. Additionally, it is the potential difference between two points that will impart one joule of energy per coulomb of charge that passes through it. It can be expressed in terms of SI base units (m, kg, s, and A)
$1\text{volte}=\frac{1\text{joule}}{\text{Coulomb}}$
View full question & answer→- 1 Joule/ Coulomb
The volt is a measure of electric potential. One volt is defined as the difference in electric potential between two points of a conducting wire when an electric current of one ampere dissipates one watt of power between those points. It is also equal to the potential difference between two parallel, infinite planes spaced 1 meter apart that create an electric field of 1 newton per coulomb. Additionally, it is the potential difference between two points that will impart one joule of energy per coulomb of charge that passes through it. It can be expressed in terms of SI base units (m, kg, s, and A)
$1\text{volte}=\frac{1\text{joule}}{\text{Coulomb}}$
Question 2461 Mark
A dielectric slab is inserted between the plates of an isolated charged capacitor. Which of the following quantities will remain the same?
Answer
When we insert a dielectric between the plates of a capacitor, induced charges of opposite polarity appear on the face of the dielectric. They build an electric field inside the dielectric, directed opposite to the original field of the capacitor.
Thus, the net effect is a reduced electric field.
Also, as the potential is proportional to the field, the potential decreases and so does the stored energy U, which is given by
$\text{U}=\frac{\text{qV}}{2}$
Thus, only the charge on the capacitor remains unchanged, as the charge is conserved in an isolated system.
View full question & answer→- The charge on the capacitor.
When we insert a dielectric between the plates of a capacitor, induced charges of opposite polarity appear on the face of the dielectric. They build an electric field inside the dielectric, directed opposite to the original field of the capacitor.
Thus, the net effect is a reduced electric field.
Also, as the potential is proportional to the field, the potential decreases and so does the stored energy U, which is given by
$\text{U}=\frac{\text{qV}}{2}$
Thus, only the charge on the capacitor remains unchanged, as the charge is conserved in an isolated system.
Question 2471 Mark
A dielectric slab is inserted between the plates of an isolated capacitor. The force between the plates will:
Answer
The force between the plates is given by
$\text{F}=\frac{\text{q}^2}{2\epsilon_0\text{A}}$
Since the capacitor is isolated, the charge on the plates remains constant.
We know that the charge is conserved in an isolated system.
Thus, the force acting between the plates remains unchanged.
View full question & answer→- Remain unchanged.
The force between the plates is given by
$\text{F}=\frac{\text{q}^2}{2\epsilon_0\text{A}}$
Since the capacitor is isolated, the charge on the plates remains constant.
We know that the charge is conserved in an isolated system.
Thus, the force acting between the plates remains unchanged.
Question 2481 Mark
How does the potential difference change with the effect of the dielectric when the battery is kept disconnected from the capacitor?
Answer
When the dielectric slab is introduced between the plates, the induced surface charge on the dielectric reduces the electric field.
The reduction in the electric field results in a decrease in potential difference.
$\text{V}=\text{Ed}=\frac{\text{E}_0\text{d}}{\text{k}}=\frac{\text{V}_0}{\text{k}}$
View full question & answer→- Decreases
When the dielectric slab is introduced between the plates, the induced surface charge on the dielectric reduces the electric field.
The reduction in the electric field results in a decrease in potential difference.
$\text{V}=\text{Ed}=\frac{\text{E}_0\text{d}}{\text{k}}=\frac{\text{V}_0}{\text{k}}$
Question 2491 Mark
"The work per unit of charge required to move a charge from a reference point to a specified point, measured in joules per coulomb or volts. The static electric field is the negative of the gradient of the electric potential." comments are given below, select the correct one:
Answer
The given statement is correct. Volt or Joules per Coulomb are SI unit of Potential (V) i.e, work done per unit charge to move a charge from a reference point to a certain point.
And electric field E is negative of the gradient of electric potential (V) i.e,$\text{E}=\frac{\text{-dv}}{\text{dr}}$
View full question & answer→- Statement is correct
The given statement is correct. Volt or Joules per Coulomb are SI unit of Potential (V) i.e, work done per unit charge to move a charge from a reference point to a certain point.
And electric field E is negative of the gradient of electric potential (V) i.e,$\text{E}=\frac{\text{-dv}}{\text{dr}}$
Question 2501 Mark
In a region of constant potential:
Answer
or we can write $|\text{E}|=-\frac{\Delta\text{V}}{\Delta\text{r}}$
The electric field intensity E and electric potential V are related as E = 0 and for V = constant, $\frac{\text{dV}}{\text{dr}}=0$ this imply that electric field intensity E = 0.
If some charge is present inside the region then electric field cannot be zero at that region, for this V = constant is not valid.
View full question & answer→- The electric field is zero.
- There can be no charge inside the region.
or we can write $|\text{E}|=-\frac{\Delta\text{V}}{\Delta\text{r}}$
The electric field intensity E and electric potential V are related as E = 0 and for V = constant, $\frac{\text{dV}}{\text{dr}}=0$ this imply that electric field intensity E = 0.
If some charge is present inside the region then electric field cannot be zero at that region, for this V = constant is not valid.