An electric charge \$10^{-3}\mu C\$ is placed at the origin (0, 0) of X-Y co-ordinate system. Two points A and B are situated at \$(\sqrt 2, \sqrt 2)\$ and (2, 0) respectively. The potential difference between the points A and B will be :

The distance of point A from origin is \$r_A=\sqrt{(\sqrt 2-0)^2+(\sqrt 2-0)^2}=\sqrt{4}=2\$The distance of point B from origin is \$r_B=\sqrt{(2-0)^2+(0-0)^2}=\sqrt{4}=2\$Potential at A due to charge \$Q=10^{-3} \mu C\$ is \$\displaystyle V_A=\frac{Q}{4\pi\epsilon_0 r_A}=\frac{Q}{8\pi\epsilon_0 }\$ as \$r_A=2\$and potential at B is \$\displaystyle V_B=\frac{Q}{4\pi\epsilon_0 r_B}=\frac{Q}{8\pi\epsilon_0 }\$ as \$r_B=2\$Potential difference between A and B \$\displaystyle=V_A-V_B=\frac{Q}{8\pi\epsilon_0 }-\frac{Q}{8\pi\epsilon_0 }=0\$ volt.

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Potential Due to a Point Charge

Q: Derive an expression for the electric potential due to a point charge :

Consider the electric potential due to a point charge q, As we move from point A, at distance rA from the charge q, to point B, at distance rB from the charge q, the change in electric potential is\$\Delta V_{BA}= V_B-V_A=- \int_A^B E. ds\$\$E.ds=[k\dfrac{q}{r^2}]\hat{r}.ds\$\$\hat{r}.ds=dr\$Only the radial distance r determines the work done or the potential. We can move through any angle we like and, as long as the radial distance remains constant, no work is done or there is no change in the electric potential.\$\Delta V_{BA}= V_B-V_A = - \int_{r_A}^{r_B}Edr=-\int_{r_A}^{r_B}[k\dfrac{q}{r^2}dr]\$\$\Delta V_{BA}= V_B-V_A=-kq[(-1)r^{-1}]_{r_A} ^{r_B}\$\$\Delta V_{BA}= V_B-V_A= kq [\dfrac{1}{r_B}-\dfrac{1}{r_B}]\$This is the change in electric potential due to a point charge as we move from rA to rB.We could ask about the change in electric potential energy as we move a charge q' from radius rA to rB due to a point charge q\$\Delta U_{BA}= kq'q[\dfrac{1}{r_B}-\dfrac{1}{r_A}]\$As with gravitational potential energy, it is more convenient -- and, therefore, useful -- to talk about the electric potential energy or the electric potential relative to some reference point. We will choose that reference point to be infinity. That is,\$r_A=\infty\$That means we can then write the electric potential at some radius r as \$V=kq\dfrac{1}{r}\$

Q: The electric potential at a point \$P(x,y,z)\$ is given by \$V = -x^2y -xz^3+4\$. The electric field \$\vec E\$ at that point is

\$V=-x^2y-xz^3+4\$\$\vec{E}=-\dfrac{\partial V}{\partial x}\hat{i}-\dfrac{\partial V}{\partial y}\hat {j}-\dfrac{\partial V}{\partial z}\hat{k}\$Now, \$\dfrac{\partial V}{\partial x}=-2xy-z^3\$\$\dfrac{\partial V}{\partial y}=-x^2\$\$\dfrac{\partial V}{\partial z}=-3xz^2\$Hence, \$\vec{E}=(2xy+z^3)\hat{i}+x^2\hat{j}+3xz^2\hat{k}\$Answer-(A)

Q: The electric potential at a point in free space due to a charge \$Q\$ coulomb is \$Q\times { 10 }^{ 11 }V\$. The electric field at that point is

Here \$V=\cfrac { Q }{ 4\pi { \varepsilon }_{ 0 }r } =Q\times { 10 }^{ 11 }\$\$\therefore 4\pi { \varepsilon }_{ 0 }r={ 10 }^{ -11 }...(i)\$Now, \$E=\cfrac { Q }{ 4\pi { \varepsilon }_{ 0 }{ r }^{ 2 } } =\cfrac { Q\times 4\pi { \varepsilon }_{ 0 } }{ (4\pi { \varepsilon }_{ 0 }{ r) }^{ 2 } } =\cfrac { Q\times 4\pi { \varepsilon }_{ 0 } }{ ({ 10 }^{ -11 })^{ 2 } } =4\pi { \varepsilon }_{ 0 }Q\times { 10 }^{ 22 }V{ m }^{ -1 }\$ (Using (i))

Q: Three charges \$-q,Q \ and \ -q\$ are placed at equal distances on a straight line. If the total potential energy of the system of three charges is zero, then the ratio \$Q:q\$ is :

Let the charges be placed as shown in the figure.The potential energy of the system is given by \$U = U_{12} + U_{13} + U_{23}\$Here, \$1,2,3\$ denote the three charges in order respectively.Let \$\displaystyle \frac{1}{4\pi\epsilon_0} = k\$Then, \$U_{12} = \displaystyle k\frac{-qQ}{d}\$, \$U_{13} = k\displaystyle\frac{q^2}{2d}\$ and \$\displaystyle U_{23} = k\frac{-qQ}{d}\$Thus, the potential energy of the system is \$U = \displaystyle \frac{k}{2d}\big(-2qQ+q^2-2qQ\big) = \frac{k}{2d}\big(q^2-4qQ\big)\$But, given that \$U = 0 \Rightarrow q^2 = 4qQ\Rightarrow Q:q = 1:4\$

Q: Two points charges \$4 \mu C\$ and \$-2 \mu C\$ are separated by a distance of 1 m in air. At what point in between the charges and on the line joining the charges, is the electric potential zero?

Let potential is zero at a distance \$x\$ from \$4\mu c\$ \$ \Rightarrow \dfrac{{k\left( 4 \right)}}{x} + \dfrac{{k\left( { - 2} \right)}}{{\left( {1 - x} \right)}} = 0\$\$ \Rightarrow \frac{4}{x} = \dfrac{2}{{\left( {1 - x} \right)}}\$\$ \Rightarrow 2 - 2x = x\$\$ \Rightarrow 2 = 3x\$\$ \Rightarrow x = \dfrac{2}{3}\$\$\therefore \$ from the distance is \$1 - \dfrac{2}{3} = \dfrac{1}{3}\$potential is zero at \$\dfrac{1}{3}m\$ from \$ - 2\mu c.\$Hence,option \$(C)\$ is correct answer.

Q: Define electric potential due to a point charge and arrive at the expression for the electric potential at a point due to a point charge in its vicinity.

The electrostatic potential at any point in an electrostatic field is defined as the work done in carrying a unit positive charge from infinity to that point against the electrostatic force of the field.Consider a point Charge \$+Q\$ at \$O\$.Let \$P\$ be a point at a distance \$r\$ from \$O\$.Consider a point \$A\$ at a distance \$x\$ from \$O\$.The magnitude of electrostatic force on unit-positive charge at \$A\$, is\$\quad F=\cfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \cfrac { Q\times i }{ { x }^{ 2 } } \$\$F=\cfrac { Q }{ 4\pi { \varepsilon }_{ 0 }{ x }^{ 2 } } \$ along \$OP\$Work Done in moving a unit positive charge from \$A\$ to \$B\$ through a small distance \$dx\$ is,\$\Delta W=\vec { F } .d\vec { x } =Fdx\cos { { 180 }^{ o } } \$\$\Delta W=-Fdx\$\$\Delta W=\cfrac { -Q }{ 4\pi { \varepsilon }_{ 0 }{ x }^{ 2 } } dx\$Total work done in moving a unit positive change from infinity to the point \$P\$ is,\$W=\displaystyle \int _{ \infty }^{ r }{ \cfrac { -1 }{ 4\pi { \varepsilon }_{ 0 } } \cfrac { Q }{ { x }^{ 2 } } } dx=-\cfrac { Q }{ 4\pi { \varepsilon }_{ 0 }{ x }^{ 2 } } \int _{ \infty }^{ r }{ \cfrac { 1 }{ x^2 } } dx\$\$W=-\cfrac { Q }{ 4\pi { \varepsilon }_{ 0 } } { \left[ \cfrac { 1 }{ x } \right] }_{ \infty }^{ r }=\cfrac { Q }{ 4\pi { \varepsilon }_{ 0 } } \left[ \cfrac { 1 }{ r } -\cfrac { 1 }{ \infty } \right] \$\$W=\cfrac { Q }{ 4\pi { \varepsilon }_{ 0 }r }\$By definition, this work done is equal to the electrostatic potential (V) at that point due to \$Q\$.\$V=\cfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \cfrac { Q }{ r } \$

Q: A charge \$3\$ coulomb experiences a force \$3000\ N\$ when placed in a uniform electric field. The potential difference between two points seperated by a distance of \$1\ cm\$ along the field lines is :

Electric force \$F=qE\$ given, \$F=3000 N, q=3 C\$ \$\therefore E=\dfrac{F}{q}=\dfrac{3000}{3}=1000 V/m\$ Potential difference, \$V={E}{d}\$ where \$d=0.01 m\$ \$\therefore V=1000\times 0.01=10 V\$ Ans:(A)

Q: A hollow metal sphere of radius \$10\$cm is charged such that the potential on its surface becomes \$80\$V. The potential at the centre of the sphere is?

The potential at the centre of the sphere is \$80\$V because it remains same at each point under the metallic hollow sphere.

Q: If \$100\$ joule of work must be done to move electric charge equal to \$4\ C\$ from a place, where potential is \$-10\$ volt to another place, where potential is V volt, find the value of V.

Given work done = 100Jcharge q = 4Cfrom potential \$V_1\$ = -10 Volts to \$V_2\$ = V Voltsso,\$W=q \triangle{V}\$\$W=q (V_2-V_1)\$\$W=4[V_2 - (-10)]\$ \$\cfrac{100}{4} = V_2 + 10\$\$V_2 = 15 \; volts\$

Physics

formula

Electric potential due to a point charge

V = \frac{Q}{4 π ϵ _{0} r}

where

Q

is a point charge and

V

is the potential due it at a point of r distance.

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Potential due to Point Charge

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An electric charge $1 0^{- 3} μ C$ is placed at the origin (0, 0) of X-Y co-ordinate system. Two points A and B are situated at $(2, 2)$ and (2, 0) respectively. The potential difference between the points A and B will be :

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Electric potential due to a point charge

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Potential Due to a Point Charge

Derive an expression for the electric potential due to a point charge :

The electric potential at a point P(x,y,z) is given by V=−x2y−xz3+4. The electric field E at that point is

The electric potential at a point in free space due to a charge Q coulomb is Q×1011V. The electric field at that point is

Three charges −q,Q and −q are placed at equal distances on a straight line. If the total potential energy of the system of three charges is zero, then the ratio Q:q is :

Two points charges 4μC and −2μC are separated by a distance of 1 m in air. At what point in between the charges and on the line joining the charges, is the electric potential zero?

Define electric potential due to a point charge and arrive at the expression for the electric potential at a point due to a point charge in its vicinity.

A charge 3 coulomb experiences a force 3000 N when placed in a uniform electric field. The potential difference between two points seperated by a distance of 1 cm along the field lines is :

A hollow metal sphere of radius 10cm is charged such that the potential on its surface becomes 80V. The potential at the centre of the sphere is?

If 100 joule of work must be done to move electric charge equal to 4 C from a place, where potential is −10 volt to another place, where potential is V volt, find the value of V.

An electric charge 10−3μC is placed at the origin (0, 0) of X-Y co-ordinate system. Two points A and B are situated at (2​,2​) and (2, 0) respectively. The potential difference between the points A and B will be :

The electric potential at a point $P (x, y, z)$ is given by $V = - x^{2} y - x z^{3} + 4$ . The electric field $E$ at that point is

The electric potential at a point in free space due to a charge $Q$ coulomb is $Q \times 10^{11} V$ . The electric field at that point is

Three charges $- q, Q a n d - q$ are placed at equal distances on a straight line. If the total potential energy of the system of three charges is zero, then the ratio $Q : q$ is :

Two points charges $4 μ C$ and $- 2 μ C$ are separated by a distance of 1 m in air. At what point in between the charges and on the line joining the charges, is the electric potential zero?

A charge $3$ coulomb experiences a force $3000 N$ when placed in a uniform electric field. The potential difference between two points seperated by a distance of $1 c m$ along the field lines is :

A hollow metal sphere of radius $10$ cm is charged such that the potential on its surface becomes $80$ V. The potential at the centre of the sphere is?

If $100$ joule of work must be done to move electric charge equal to $4 C$ from a place, where potential is $- 10$ volt to another place, where potential is V volt, find the value of V.

An electric charge $1 0^{- 3} μ C$ is placed at the origin (0, 0) of X-Y co-ordinate system. Two points A and B are situated at $(2, 2)$ and (2, 0) respectively. The potential difference between the points A and B will be :