Electric Potential and Potential Difference Questions in English

(N/A) source of electricity,such as a cell,battery,or power supply,helps to maintain a potential difference across a conductor. This potential difference is necessary to drive the flow of electrons through the conductor,thereby creating an electric current.

(N/A) The potential difference between two points is said to be $1$ $V$ if $1$ $J$ of work is done in moving a charge of $1$ $C$ from one point to the other.
Mathematically,$V = W / Q$,where $V = 1$ $V$,$W = 1$ $J$,and $Q = 1$ $C$.

(B) The energy given to each coulomb of charge is equal to the amount of work required to move it.
The amount of work done is calculated using the formula:
Potential difference $(V)$ = Work done $(W)$ / Charge $(Q)$
Work done $(W)$ = Potential difference $(V)$ $\times$ Charge $(Q)$
Given:
Charge $(Q)$ = $1\, C$
Potential difference $(V)$ = $6\, V$
Calculation:
Work done $(W)$ = $6\, V \times 1\, C = 6\, J$
Therefore,$6\, J$ of energy is given to each coulomb of charge passing through a battery of $6\, V$.

(B) To measure the potential difference between two points in an electrical circuit,a voltmeter must be connected in parallel across those two points.
This is because a voltmeter has a very high resistance,which ensures that it does not draw significant current from the circuit,thereby maintaining the accuracy of the potential difference measurement.

(A) Voltage $(V)$ is defined as the work done $(W)$ per unit charge $(Q)$ to move a charge between two points.
Mathematically,$V = \frac{W}{Q}$.
We know that electric current $(I)$ is the rate of flow of charge,given by $I = \frac{Q}{t}$,which implies $Q = I \times t$.
Substituting this value of $Q$ into the voltage formula,we get $V = \frac{W}{I \times t}$.
Therefore,voltage is represented by $\frac{\text{Work}}{\text{Current} \times \text{Time}}$.

(N/A) The electrostatic potential at a point in an electric field is defined as the amount of work done in moving a unit positive charge from infinity to that specific point against the electrostatic force.

(B) The $SI$ unit of electric potential is the volt $(V)$.
It is defined as the amount of work done in moving a unit positive charge from infinity to a point in an electric field.

(N/A) The electric potential $(V)$ at a point in an electric field is defined as the amount of work $(W)$ done in moving a unit positive charge $(q)$ from infinity to that point.
Mathematically,it is expressed as:
$V = \frac{W}{q}$
Where:
$V$ = Electric potential
$W$ = Work done
$q$ = Charge

(A) Electrostatic potential is defined as the amount of work done in moving a unit positive charge from infinity to a specific point in an electric field.
Since work is a scalar quantity and charge is a scalar quantity,their ratio,which defines the potential,is also a scalar quantity.
Therefore,electrostatic potential is a scalar quantity.

(N/A) Potential difference between two points in an electric circuit is defined as the amount of work done in moving a unit positive charge from one point to the other.
It is a scalar quantity because it only has magnitude and no specific direction.
The $SI$ unit of potential difference is the volt $(V)$,which is equivalent to joules per coulomb $(J/C)$.

(A) The potential difference between two points is defined as the work done per unit charge in moving a charge from one point to another.
Mathematically,$V = W / Q$.
If $W = 1 \text{ joule}$ and $Q = 1 \text{ coulomb}$,then $V = 1 \text{ joule} / 1 \text{ coulomb} = 1 \text{ volt}$.
Therefore,the potential difference between two points is said to be $1$ volt if $1$ joule of work is done to move a charge of $1$ coulomb between those two points.

(N/A) To maintain a constant potential difference across the ends of a conductor,we need to connect it to a source of electrical energy,such as a cell or a battery.
This source of electromotive force $(emf)$ performs work to move charges,thereby sustaining the flow of current through the conductor.

(A) The potential difference between two points is defined as the work done in moving a unit positive charge from one point to another. The $SI$ unit of potential difference is the volt $(V)$,where $1 \text{ volt} = \frac{1 \text{ joule}}{1 \text{ coulomb}}$. Therefore,the single word for $\frac{1 \text{ joule}}{1 \text{ coulomb}}$ is $1 \text{ volt}$.

(N/A) The potential difference across the ends of a conductor is measured using a device called a voltmeter.
$A$ voltmeter is always connected in parallel across the points between which the potential difference is to be measured.
It has a high resistance to ensure that it draws a negligible amount of current from the circuit.

(B) voltmeter is a device used to measure the potential difference between two points in an electrical circuit.
To measure this potential difference accurately without significantly altering the current flowing through the circuit,it must have a very high resistance.
Therefore,it is always connected in parallel across the component or the points where the potential difference needs to be measured.

(N/A) The device used to measure the potential difference between two points in an electric circuit is called a $Voltmeter$.
To measure the potential difference, a $Voltmeter$ is always connected in $\parallel$ across the two points where the potential difference is to be measured.

(N/A) Potential: The electric potential at a point in an electric field is defined as the amount of work done in moving a unit positive charge from infinity to that point.
Potential difference: The potential difference between two points in an electric circuit is defined as the amount of work done in moving a unit positive charge from one point to the other.
The $SI$ unit for both electric potential and potential difference is the volt $(V)$.

(N/A) The potential difference between two points in a current-carrying conductor is said to be $1$ volt when $1$ joule of work is done to move a charge of $1$ coulomb from one point to the other.
Mathematically,$V = W / Q$,where $V = 1$ volt,$W = 1$ joule,and $Q = 1$ coulomb.

(N/A) device that maintains a potential difference across a conductor is called a cell or a battery.
The circuit symbol for a cell consists of two parallel vertical lines,where the longer line represents the positive terminal $(+)$ and the shorter,thicker line represents the negative terminal $(-)$.
Circuit symbol for a cell:
$| |$ (Longer line: $+$,Shorter line: $-$)

(N/A) voltmeter is an electrical instrument used to measure the potential difference between two points in an electrical circuit.
It is always connected in parallel across the component or the part of the circuit where the potential difference is to be measured.

(A) The electric potential $V$ at a point is defined as the amount of work done $W$ in moving a unit positive charge $q$ from infinity to that point.
The formula is given by $V = \frac{W}{q}$.
Given values are $W = 2 \times 10^{-3} \text{ J}$ and $q = 10 \times 10^{-6} \text{ C}$.
Substituting these values into the formula:
$V = \frac{2 \times 10^{-3}}{10 \times 10^{-6}} = \frac{2}{10} \times 10^{(-3 + 6)} = 0.2 \times 10^3 = 200 \text{ V}$.
Therefore, the potential at point $P$ is $200 \text{ V}$.

(N/A) In metals,free electrons are in a state of random thermal motion. In the absence of an external electric field,the average velocity of these electrons in any given direction is zero,meaning there is no net flow of charge. Current is defined as the rate of flow of charge. For a net flow to occur,an external electric potential difference must be applied across the conductor. This potential difference creates an electric field that exerts a force on the electrons,causing them to drift in a specific direction,thereby establishing an electric current.

(N/A) Charges can move in a conductor only if there is a potential difference across its two ends. This potential difference is necessary to create an electric field that exerts a force on the charges,causing them to flow. The device used to maintain this potential difference in an electric circuit is an electric cell or a battery (a combination of one or more cells).

(A) Given: Potential difference $(V) = 20 \ V$,Work $(W) = 40 \ J$.
We know that the work done in moving a charge $(Q)$ between two points is given by the formula:
$W = Q \times V$
Substituting the given values:
$40 \ J = Q \times 20 \ V$
$Q = \frac{40 \ J}{20 \ V} = 2 \ C$
We also know that the total charge $(Q)$ is related to the number of electrons $(n)$ by the relation:
$Q = n \times e$
where $e$ is the elementary charge,$e = 1.6 \times 10^{-19} \ C$.
Therefore,the number of electrons $(n)$ is:
$n = \frac{Q}{e} = \frac{2 \ C}{1.6 \times 10^{-19} \ C} = 1.25 \times 10^{19}$

(N/A) One volt is defined as the potential difference between two points in a current-carrying conductor when $1 \ J$ of work is done to move a charge of $1 \ C$ from one point to the other.
$(b)$ The relation between potential difference $(V)$,work $(W)$,and charge $(Q)$ is given by the formula: $V = W / Q$.
Given:
Work $(W)$ = $100 \ J$
Charge $(Q)$ = $20 \ C$
Using the formula $V = W / Q$:
$V = 100 \ J / 20 \ C = 5 \ V$.
Therefore,the potential difference between the two terminals of the battery is $5 \ V$.

$(a)$ Electric potential difference between two points in an electric circuit carrying current is defined as the work done to move a unit positive charge from one point to the other.
$(b)$ The instrument used to measure potential difference is a Voltmeter. It is always connected in parallel across the points between which the potential difference is to be measured.
$(c)$ Given: Current $(I) = 2 \, A$, Time $(t) = 1 \, \text{minute} = 60 \, s$, Potential difference $(V) = 3 \, V$.
First, calculate the total charge $(Q)$ transferred: $Q = I \times t = 2 \, A \times 60 \, s = 120 \, C$.
Now, calculate the work done $(W)$ using the formula $W = V \times Q$: $W = 3 \, V \times 120 \, C = 360 \, J$.
Thus, the work done is $360 \, J$.

(B) Given: Charge $(Q) = 2 \, C$,Potential difference $(V) = 12 \, V$.
The formula for work done $(W)$ in moving a charge between two points is given by $W = Q \times V$.
Substituting the given values into the formula:
$W = 2 \, C \times 12 \, V = 24 \, J$.
Therefore,the work done is $24 \, J$.

(B) Given: Charge $q = 2 \ C$,Initial potential $V_1 = 118 \ V$,Final potential $V_2 = 128 \ V$.
We know that the potential difference $V$ is defined as the work done $W$ per unit charge $q$,i.e.,$V = W / q$.
Therefore,the work done $W = q \times (V_2 - V_1)$.
Substituting the values: $W = 2 \times (128 - 118) = 2 \times 10 = 20 \ J$.
Thus,the total work done is $20 \ J$.

(D) Electric current is defined as the flow of electric charge from a region of higher potential to a region of lower potential.
When two charged bodies are connected,the flow of charge occurs due to a potential difference between them.
If both bodies have the same potential,there is no potential difference $(V_1 - V_2 = 0)$.
Consequently,no current will flow between the two bodies.

(B) Electric potential and potential difference are scalar quantities.
They are defined as the amount of work done in moving a unit positive charge from one point to another.
Since they only have magnitude and no specific direction associated with them,they are classified as scalars.

(A) The electric potential at a point is defined as the amount of work done in moving a unit positive charge from infinity to that point.
If a body is positively charged,it exerts a repulsive force on a unit positive charge,meaning work must be done against the electric field to bring the charge closer. Thus,the potential is positive.
Conversely,if a body is negatively charged,it exerts an attractive force on a unit positive charge,meaning the field does the work. Thus,the potential is negative.
Therefore,the statement is True.

(A) By convention,the direction of electric current is taken as the direction of flow of positive charges. Positive charges always move from a region of higher electric potential to a region of lower electric potential. This flow continues until the potential difference between the two points becomes zero.

(A) battery or a cell is a device that converts chemical energy into electrical energy. It maintains a constant potential difference (voltage) across the ends of a conductor,which drives the flow of electrons (current) through the circuit. Therefore,the statement is True.

(A) Voltage $(V)$ is defined as the work done $(W)$ per unit charge $(Q)$ to move it between two points in an electric circuit.
Mathematically,$V = W / Q$.
Since electric current $(I)$ is defined as the rate of flow of charge,$I = Q / t$,which implies $Q = I \times t$.
Substituting $Q = I \times t$ into the voltage formula,we get $V = W / (I \times t)$.
Therefore,the correct formula is $\text{Work} / (\text{Current} \times \text{Time})$.

(B) Electric potential difference $(V)$ is defined as the work done $(W)$ per unit charge $(Q)$ in moving a positive charge from one point to another.
Mathematically,$V = W / Q$.
The unit of work $(W)$ is Joule $(J)$ and the unit of charge $(Q)$ is Coulomb $(C)$.
Therefore,the unit of electric potential difference is $J/C$,which is also known as the Volt $(V)$.

(C) The potential difference $(V)$ between two points is defined as the work done $(W)$ per unit charge $(Q)$ to move the charge from one point to another.
Formula: $V = \frac{W}{Q}$
Given:
Work done $(W)$ = $15 \,J$
Charge $(Q)$ = $3 \,C$
Calculation:
$V = \frac{15 \,J}{3 \,C} = 5 \,V$
Therefore,the potential difference between the two points is $5 \,V$.

(C) The electric potential difference between two points in an electric circuit is defined as the amount of work done in moving a unit charge from one point to the other.
Its $SI$ unit is the volt $(V)$,named after the Italian physicist Alessandro Volta.
$1 \text{ volt} = 1 \text{ joule} / 1 \text{ coulomb}$.
Therefore,the correct unit is the volt.

(B) The electric potential difference $(V)$ between two points is defined as the amount of work done $(W)$ in moving a unit positive charge $(Q)$ from one point to another.
Mathematically,$V = W / Q$.
The unit of work $(W)$ is Joule $(J)$ and the unit of charge $(Q)$ is Coulomb $(C)$.
Therefore,the unit of electric potential difference is $Joule/Coulomb$ $(J/C)$,which is also known as Volt $(V)$.

(C) The $SI$ unit of electric potential (or potential difference) is $volt$ $(V)$.
Electric current is measured in $ampere$ $(A)$.
Resistance is measured in $ohm$ ($\Omega$).
Electric energy is measured in $joule$ $(J)$ or $kilowatt-hour$ $(kWh)$.
Therefore, the correct option is $C$.

(B) The device used to measure the electric potential difference between two points in an electric circuit is called a $Voltmeter$.
It is always connected in parallel across the points where the potential difference is to be measured.
An $Ammeter$ is used to measure electric current.
$A$ $Galvanometer$ is used to detect the presence of small electric currents.
$A$ $Voltameter$ is an electrolytic cell used for measuring the quantity of electricity by the amount of chemical action produced.

(A) The work done $(W)$ in moving a charge $(Q)$ through an electric potential difference $(V)$ is given by the formula: $W = Q \times V$.
Given:
Charge $(Q)$ = $2 \, C$
Potential difference $(V)$ = $6 \, V$
Substituting the values into the formula:
$W = 2 \, C \times 6 \, V = 12 \, J$.
Therefore,the work done is $12 \, J$.