Q: How much work is done in moving a charge of \$2\ C\$ across two points having a potential difference of \$12\ V?\$

Find work done \$=W\$Given,Moving charge \$=2C=(Q)\$Two points having a potential difference of \$=12V\$\$h=QV\\ \quad=2\times12\\ \quad=24J\$

Q: In the circuit shown the current through the \$4\Omega\$ resistor is \$1\$ amp when the points \$P\$ and \$M\$ are connected to a d.c voltage source. The potential difference between the points \$M\$ and \$N\$ is:

In the given fig\$:-\$In parallel resistances\$,\$ the potential differences across them is same\$.\$potential difference across \$PM\$ \$V = 4 \times 1 = 4\,volt\,\,\left( {i.e,\,across\,4\Omega } \right)\$Equivalent resistance of lower side arm\$,\$\${R_{eq}} = 1 + \frac{{0.5 \times 0.5}}{{0.5 + 0.5}} = 1.25\Omega \$Let current \$I\$ flow in lower branch\$,\$So\$,\$ \$1.25I = 4V\$\$I = \frac{4}{{1.25}} = 3.2A\$therefore\$,\$\$3.2A\$ currents flows in \$1\Omega \$ resistance\$,\$ hence potential difference between \$M\$ and \$N\$ is \$V' = 3.2 \times 1 = 3.2\,volt\$ Hence,option \$(B)\$ is correct answer.

Q: Two positive point charges are of \$12C\$ and \$8C\$ are \$10cm\$ apart from each other. The work done in bringing them \$4cm\$ closer is

Given two charges \$q_{1}=12\mu C\$, \$q_{2}=8\mu C\$ are at distance '\$d_{1}\$'\$=10\, cm \$We have to find the work done in bringing them \$d_{2}=4\, cm\$ closer.For this, we find the charge in the potential energy of the system and then take the difference.Now \$d_{1}=10\, cm=10\times 10^{-2}=10^{-1}\, m\$\$d_{2}=4\, cm=4\times 10^{-2}\, m\$So, potential (\$V_{1}\$) when \$q_{1}\$ and \$q_{2}\$ are distance \$d_{1}\$ apart, \$V_{1}=\dfrac{1}{4\pi \varepsilon _{0}}\dfrac{q_{1}q_{2}}{d_{1}}\$Potential (\$V_{2}\$) when \$q_{1}\$ and \$q_{2}\$ are distance \$d_{2}\$ apart, \$V_{2}=\dfrac{1}{4\pi \varepsilon _{0}}\dfrac{q_{1}q_{2}}{d_{2}}\$Work done \$=V_{2}-V_{1}=\dfrac{1}{4\pi \varepsilon _{0}}q_{1}q_{2}\left ( \dfrac{1}{d_{2}}-\dfrac{1}{d_{1}} \right )\$\$=9\times 10^{9}\times 12\times 8\times 10^{-12}\left ( \dfrac{1}{4\times 10^{-2}}-\dfrac{1}{10\times 10^{-2}} \right )\$\$=864\times 10^{-3}\left [ \dfrac{6\times 10^{2}}{40\times 10^{-2}} \right ]\$\$=129.6\times 10^{1}\$\$=12.96\approx 13\; Joules\$

Q: \$27\$ drops of the same size are charged at \$220\ V\$ each. They combine to form a bigger drop. Calculate the potential of the bigger drop.

The radius of the bigger drop will be \$\dfrac{4}{3} \pi R^3 = 27 \times \dfrac{4}{3} \pi r^3\$\$R = 3r \$the potential of the small drop = \$\dfrac{KQ}{r} \$ =\$ v \$The potential of the big drop \$Q = nq = 27 q\$\$V' = \dfrac{KQ}{R}\$ = \$\dfrac{K \times 27q }{3r}\$ = \$ 9v\$ \$9 \times 220\$ = \$1980 \ V\$

Q: Two points P and Q are maintained at potentials of of \$10V\$ and \$-4V\$ respectively. The work done in moving \$100\$ electrons from P to Q is equal to

potential difference=\$V=-4-10=-14V\$Charge=\$q=100e=-1.6\times 10^{-17}C\$Now, potential difference, \$V=\dfrac{W}{q}\$\$\implies W=qV=-1.6\times 10^{-17}\times (-14) \$\$\implies W=2.24\times 10^{-16}J\$Answer-(A)

Q: Figure shows some equipotential lines distributed in space. A charged object is moved from point A to point B.

In all the three figures, \${V}_{A}=20V\$ and \${V}_{B}=40V\$Work done in carrying a charge \$q\$ from \$A\$ to \$B\$ is \$W=q({ V }_{ B }-{ V }_{ A })\quad \quad \$Hence the work done is same in all figures.

Question 1

How much work is done in moving a charge of $2\ C$ across two points having a potential difference of $12\ V?$

Accepted Answer

Find work done $=W$Given,Moving charge $=2C=(Q)$Two points having a potential difference of $=12V$$h=QV\ \quad=2\times12\ \quad=24J$

Question 2

In the circuit shown the current through the $4\Omega$ resistor is $1$ amp when the points $P$ and $M$ are connected to a d.c voltage source. The potential difference between the points $M$ and $N$ is:

Accepted Answer

In the given fig$:-$In parallel resistances$,$ the potential differences across them is same$.$potential difference across $PM$  $V = 4 \times 1 = 4\,volt\,\,\left( {i.e,\,across\,4\Omega } \right)$Equivalent resistance of lower side arm$,$${R_{eq}} = 1 + \frac{{0.5 \times 0.5}}{{0.5 + 0.5}} = 1.25\Omega $Let current $I$ flow in lower branch$,$So$,$ $1.25I = 4V$$I = \frac{4}{{1.25}} = 3.2A$therefore$,$$3.2A$ currents flows in $1\Omega $ resistance$,$ hence potential difference between $M$ and $N$ is $V' = 3.2 \times 1 = 3.2\,volt$ Hence,option $(B)$ is correct answer.

Question 3

Charges $+q$ and $-q$ are placed at points $A$ and $B$ respectively which are a distance $2L$ apart, $C$ is the midpoint between $A$ and $B$. The work done in moving a charge $+Q$ along the semicircle $CRD$ is:

Accepted Answer

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Question 4

Two positive point charges are of $12C$ and $8C$ are $10cm$ apart from each other. The work done in bringing them $4cm$ closer is

Accepted Answer

Given two charges $q_{1}=12\mu C$, $q_{2}=8\mu C$ are at distance '$d_{1}$'$=10\, cm $We have to find the work done in bringing them $d_{2}=4\, cm$ closer.For this, we find the charge in the potential energy of the system and then take the difference.Now $d_{1}=10\, cm=10\times 10^{-2}=10^{-1}\,  m$$d_{2}=4\, cm=4\times 10^{-2}\,  m$So, potential ($V_{1}$) when $q_{1}$ and $q_{2}$ are distance $d_{1}$ apart, $V_{1}=\dfrac{1}{4\pi \varepsilon _{0}}\dfrac{q_{1}q_{2}}{d_{1}}$Potential ($V_{2}$) when $q_{1}$ and $q_{2}$ are distance $d_{2}$ apart, $V_{2}=\dfrac{1}{4\pi \varepsilon _{0}}\dfrac{q_{1}q_{2}}{d_{2}}$Work done $=V_{2}-V_{1}=\dfrac{1}{4\pi \varepsilon _{0}}q_{1}q_{2}\left ( \dfrac{1}{d_{2}}-\dfrac{1}{d_{1}} \right )$$=9\times 10^{9}\times 12\times 8\times 10^{-12}\left ( \dfrac{1}{4\times 10^{-2}}-\dfrac{1}{10\times 10^{-2}} \right )$$=864\times 10^{-3}\left [ \dfrac{6\times 10^{2}}{40\times 10^{-2}} \right ]$$=129.6\times 10^{1}$$=12.96\approx 13\; Joules$

Question 5

The work done in bringing a unit positive charge from infinite distance to a point at distance x from a positive charge Q is W. Then the potential $\phi$ at that point is:

Accepted Answer

Electric potential at a point in an electric field is defined as the work done in bringing a unit positive charge from infinity to that point. Or,   $V = \cfrac{W}{q}$Since charge is of magnitude unity, Hence electric potential at that point will be equal to the workdone $\therefore \phi =W$Option B is correct.

Question 6

$27$ drops of the same size are charged at $220\ V$ each. They combine to form a bigger drop. Calculate the potential of the bigger drop.

Accepted Answer

The radius of the bigger drop will be $\dfrac{4}{3} \pi R^3 = 27 \times \dfrac{4}{3} \pi r^3$$R = 3r  $the potential of the small drop                         = $\dfrac{KQ}{r} $ =$ v $The potential of the big drop    $Q = nq      =     27 q$$V' = \dfrac{KQ}{R}$        =   $\dfrac{K \times 27q }{3r}$             =  $ 9v$  $9 \times 220$     =     $1980 \ V$

Question 7

The dimensional formula of electric potential is given by

Accepted Answer

Electric potential is defined as the work done to move per unit positive charge from one point to another point.The SI unit of electric potential is $volt(V)$.$V=\dfrac{W}{Q}$Hence dimension is $\dfrac{[ML^2T^{-2}]}{[AT]}$$=[ML^2T^{-3}A^{-1}]$Answer-(A)

Question 8

The kinetic energy of an electron, which is accelerated in the potential difference of $100\ V$, is

Accepted Answer

eV i.e. electron volt is the unit of energy defined as the kinetic energy of an electron accelerated in the potential difference of V volt $\& $ e is charg of electron.Thus acc. to questionK.E.=$e\times 100 V$K.E.=$1.6\times {10}^{-19}\times 100 V$K.E.=$1.6\times {10}^{-17}$

Question 9

Two points P and Q are maintained at potentials of of $10V$ and $-4V$ respectively. The work done in moving $100$ electrons from P to Q is equal to

Accepted Answer

potential difference=$V=-4-10=-14V$Charge=$q=100e=-1.6\times 10^{-17}C$Now, potential difference, $V=\dfrac{W}{q}$$\implies W=qV=-1.6\times 10^{-17}\times (-14) $$\implies W=2.24\times 10^{-16}J$Answer-(A)

Question 10

Figure shows some equipotential lines distributed in space. A charged object is moved from point A to point B.

Accepted Answer

In all the three figures, ${V}_{A}=20V$ and ${V}_{B}=40V$Work done in carrying a charge $q$ from $A$ to $B$ is $W=q({ V }_{ B }-{ V }_{ A })\quad \quad $Hence the work done is same in all figures.

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Potential Difference

How much work is done in moving a charge of $2 C$ across two points having a potential difference of $12 V ?$

In the circuit shown the current through the $4 Ω$ resistor is $1$ amp when the points $P$ and $M$ are connected to a d.c voltage source. The potential difference between the points $M$ and $N$ is:

Charges $+ q$ and $- q$ are placed at points $A$ and $B$ respectively which are a distance $2 L$ apart, $C$ is the midpoint between $A$ and $B$ . The work done in moving a charge $+ Q$ along the semicircle $C R D$ is:

Two positive point charges are of $12 C$ and $8 C$ are $10 c m$ apart from each other. The work done in bringing them $4 c m$ closer is

The work done in bringing a unit positive charge from infinite distance to a point at distance x from a positive charge Q is W. Then the potential $ϕ$ at that point is:

$27$ drops of the same size are charged at $220 V$ each. They combine to form a bigger drop. Calculate the potential of the bigger drop.

The dimensional formula of electric potential is given by

The kinetic energy of an electron, which is accelerated in the potential difference of $100 V$ , is

Two points P and Q are maintained at potentials of of $10 V$ and $- 4 V$ respectively. The work done in moving $100$ electrons from P to Q is equal to

Figure shows some equipotential lines distributed in space. A charged object is moved from point A to point B.

Revise with Concepts

Advanced Knowledge of Electrostatic Potential Energy and Potential Difference

Learn with Videos

Quick Summary With Stories

Potential Difference

How much work is done in moving a charge of 2 C across two points having a potential difference of 12 V?

In the circuit shown the current through the 4Ω resistor is 1 amp when the points P and M are connected to a d.c voltage source. The potential difference between the points M and N is:

Charges +q and −q are placed at points A and B respectively which are a distance 2L apart, C is the midpoint between A and B. The work done in moving a charge +Q along the semicircle CRD is:

Two positive point charges are of 12C and 8C are 10cm apart from each other. The work done in bringing them 4cm closer is

The work done in bringing a unit positive charge from infinite distance to a point at distance x from a positive charge Q is W. Then the potential ϕ at that point is:

27 drops of the same size are charged at 220 V each. They combine to form a bigger drop. Calculate the potential of the bigger drop.

The dimensional formula of electric potential is given by

The kinetic energy of an electron, which is accelerated in the potential difference of 100 V, is

Two points P and Q are maintained at potentials of of 10V and −4V respectively. The work done in moving 100 electrons from P to Q is equal to

Figure shows some equipotential lines distributed in space. A charged object is moved from point A to point B.

How much work is done in moving a charge of $2 C$ across two points having a potential difference of $12 V ?$

In the circuit shown the current through the $4 Ω$ resistor is $1$ amp when the points $P$ and $M$ are connected to a d.c voltage source. The potential difference between the points $M$ and $N$ is:

Charges $+ q$ and $- q$ are placed at points $A$ and $B$ respectively which are a distance $2 L$ apart, $C$ is the midpoint between $A$ and $B$ . The work done in moving a charge $+ Q$ along the semicircle $C R D$ is:

Two positive point charges are of $12 C$ and $8 C$ are $10 c m$ apart from each other. The work done in bringing them $4 c m$ closer is

The work done in bringing a unit positive charge from infinite distance to a point at distance x from a positive charge Q is W. Then the potential $ϕ$ at that point is:

$27$ drops of the same size are charged at $220 V$ each. They combine to form a bigger drop. Calculate the potential of the bigger drop.

The kinetic energy of an electron, which is accelerated in the potential difference of $100 V$ , is

Two points P and Q are maintained at potentials of of $10 V$ and $- 4 V$ respectively. The work done in moving $100$ electrons from P to Q is equal to