A regular hexagon of side 10 cm has a charge \$5 \mu C\$ at each of its vertices. Calculate the potential at the centre of the hexagon.

Let \$O\$ be the centre of the hexagon It follows that the point \$O\$, when joined to the ends of a side of the hexagon forms an equilateral triangle.\$\therefore AO=BO=CO=DO=EO=FO=10cm=0.1\ m\$Since at each comer of the hexagon, a charge of \$5\$pc i.e., \$5\times 10-6C\$ is placed, total electric potential at point \$o\$ due to the charges at the six comers,\$V=6\times \left (\dfrac {1}{4\pi \in_0}\times \dfrac {q}{r}\right)\$\$=6\times 9\times 10^9 \times \dfrac {5\times 10^{-6}}{0.1}\$\$=2.7\times 10^6 V\$.

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>>Potential Due to System of Charges
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Electric potential due to system of charges

Q: Two charges \$ 5 \times 10^{-8}\ C\$ and \$-3\times 10^{-8}\ C\$ are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

Let p be the point of zero potential be p at distance r from the charge 1..\$d=16\ cm\$Electric potential, \$v = q_1/4\pi\epsilon_or+ q_2/4\pi\epsilon_o(d-r)\$for \$v=0\$, \$\displaystyle \frac{q_1}{r}=-\frac{q_2}{d-r}\$\$\Rightarrow \displaystyle \frac{5\times 10^{-8}}{r}=\frac{-3\times 10^{-8}}{0.16-r}\$\$r=10\ cm\$If, point p is outside,\$v = q_1/4\pi\epsilon_or+ q_2/4\pi\epsilon_o(r-d)\$so here, \$r=40\ cm\$

Q: Three charges \$2q, - q, -q\$ are located at the vertices of an equilateral triangle. At the centre of the triangle :

As it is equilateral triangle so the distances of each charges from center are equal i.e \$r\$ (say) Potential at center is \$V=k(2q/r-q/r-q/r)=0\$.Electric field is a vector quantity so the resultant field at the center will not be zero.

Q: Four electric charges \$+q, +q, -q\$ and \$-q\$ are placed at the corners of a square of side \$2L\$ (see figure). The electric potential at point A, midway between the two charges \$+q\$ and \$+q\$, is

Here, \$AB=AE=L\$ and \$AC=AD=\sqrt{AB^2+BC^2}=\sqrt{L^2+(2L)^2}=\sqrt{5}L\$ Potential at A is \$\displaystyle V_A=\dfrac{1}{4\pi\epsilon_0}\left[\dfrac{q}{AB}+\dfrac{-q}{AC}+\dfrac{-q}{AD}+\dfrac{q}{AE}\right]\$ or \$\displaystyle V_A=\dfrac{1}{4\pi\epsilon_0}\left[\dfrac{q}{L}+\dfrac{-q}{\sqrt{5}L}+\dfrac{-q}{\sqrt{5}L}+\dfrac{q}{L}\right]=\dfrac{1}{4\pi\epsilon_0} \dfrac{2q}{L}\left[1-\dfrac{1}{\sqrt{5}}\right]\$

Q: Two charges -q and +q are located at points (0, 0, -a) and (0, 0, a), respectively. (a) What is the electrostatic potential at the points (0, 0, z) and (x, y, 0) ?(b) Obtain the dependence of potential on the distance r of a point from the origin when r/a >> 1(c) How much work is done in moving a small test charge from the point (5,0,0) to (-7,0,0) along the x-axis? Does the answer change if the path of the test charge between the same points is not along the x-axis?

(a) Zero at both the pointsCharge − q is located at (0, 0, − a) and charge + q is located at (0, 0, a). Hence, they form a dipole. Point (0, 0, z) is on the axis of this dipole and point (x, y, 0) is normal to the axis of the dipole. Hence, electrostatic potential at point (x, y, 0) is zero. Electrostatic potential at point (0, 0, z) is given by,\$\displaystyle V=\frac{1}{4\pi\in_0}\left(\frac{q}{z-a}\right)+\frac{1}{4\pi\in_0}\left(-\frac{q}{z+a}\right)\$ \$\displaystyle =\frac{q(z+a-z+a)}{4\pi\in_0(z^2-a^2)}\$ \$\displaystyle =\frac{2qa}{4\pi\in_0(z^2-a^2)}=\frac{p}{4\pi\in_0(z^2-a^2)}\$Where,\$\in_0=\$ Permittivity of free space\$p=\$ Dipole moment of the system of two charges \$=2qa\$(b) Distance r is much greater than half of the distance between the two charges. Hence, the potential (V) at a distance r is inversely proportional to square of the distance i.e., \$V \propto\frac{1}{r^2}\$ (c) Zero. The answer does not change if the path of the test is not along the x-axis.A test charge is moved from point (5, 0, 0) to point (−7, 0, 0) along the x-axis. Electrostatic potential \$(V_1)\$ at point (5, 0, 0) is given by,Electrostatic potential, \$V_2\$, at point (− 7, 0, 0) is given by,\$\displaystyle V_2=\frac{-q}{4\pi\in_0}\frac{1}{\sqrt{(-7)^2+(-a)^2}}+\frac{q}{4\pi\in_0}\frac{1}{\sqrt{(-7)^2+(a)^2}}\$ \$=\displaystyle \frac{-q}{4\pi\in_0\sqrt{49+a^2}}+\frac{q}{4\pi\in_0}\frac{1}{\sqrt{49+a^2}}\$ \$=0\$Hence, no work is done in moving a small test charge from point (5, 0, 0) to point (−7, 0, 0) along the x-axis.The answer does not change because work done by the electrostatic field in moving a test charge between the two points is independent of the path connecting the two points.

Q: Four point charges -Q,-q,2q and 2Q are placed, one at each corner of the square. The relation between Q and q fo which the potential at the centre of the square is zero, is

Let the side of the square be \$=a\$Potential at centre\${ V }_{ c }=\dfrac { K\times -Q }{ \dfrac { a }{ \sqrt { 2 } } } +\dfrac { K\times -q }{ \dfrac { a }{ \sqrt { 2 } } } +\dfrac { K\times 2q }{ \dfrac { a }{ \sqrt { 2 } } } +\dfrac { K\times 2Q }{ \dfrac { a }{ \sqrt { 2 } } } \$\$0=\dfrac { KQ }{ \dfrac { a }{ \sqrt { 2 } } } +\dfrac { Kq }{ \dfrac { a }{ \sqrt { 2 } } } \$\$\Rightarrow Q=-q\$Option (C) is correct

Q: A cube of side b has a charge q at each of its vertices. Determine the potential and electric field due to this charge array at the centre of the cube.

Diagonal of the cube face, \$d=\sqrt(b^2+b^2)=b\sqrt2\$Diagonal of cube \$l=\sqrt(d^2+b^2)=b\sqrt3\$Distance between center of cube and vertices, \$r=l/2=b\sqrt3/2\$Potential, \$v = 8q/4\pi\epsilon_or=4q/\sqrt3\pi\epsilon_ob\$ (8 due to presence of 8 charges at the vertices)by symmetry of charge electric field, E=0.

Q: \$27\$ identical drops of mercury are charged simultaneously with the same potential of \$10\ V\$. Assuming the drop to the spherical, if all the charged drops are made to combine to form one large drop, then its potential will be _____ \$V\$.

The volume of spheres.\$27\times \dfrac {4}{3}\pi r^{3} = \dfrac {4}{3} \pi R^{3}\$where, \$r\$ and \$R\$ are the volume of smaller and bigger sphere respectively.\$\Rightarrow R = 3r\$Now, initial potential of smaller sphere,\$V_{1} = \dfrac {1}{4} \pi \epsilon_{0}\dfrac {Q}{r}\$\$\therefore Q = V_{1}r\times 4\pi \epsilon_{0}\$\$\therefore\$ Charge on \$27\$ spheres, we have\$Q_{T} = 27\times 10\times r \times 4\pi \epsilon_{0}\$Now, charge on bigger sphere\$Q_{B} = V_{2}R \times 4\pi \epsilon_{0}\$Now, \$Q_{T} = Q_{B}\$\$\Rightarrow 270\times r \times 4\pi \epsilon_{0} = V_{2}R\times 4\pi \epsilon_{0}\$\$\Rightarrow 270r = V_{2}\times 3r\Rightarrow V_{2} = 90\ V\$.

Q: Point charge \$+4q\$, \$-q\$ and \$+4q\$ are kept on the x-axis at points \$x=0, x=a\$ and \$x=4a\$ respectively. Then:

At the present position all the charges are in equilibrium. But when they displaced slightly from their present position, they do not return back. So, they are in unstable equilibrium position.

Q: Calculate the electric potential at the center of the square :

Let the distance between each corner and the center of square be \$x\$.\$\therefore\$ \$s^2 =x^2 + x^2 \$ \$\implies x = \dfrac{s}{\sqrt{2}}\$Potential at the center due to charge at corner 1, \$V_1 = \dfrac{k q}{x} = \dfrac{\sqrt{2}kq}{s}\$The square is symmetric about its center.Thus total potential at the center \$V =4V_1 = k\dfrac{4\sqrt{2} q}{s}\$

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Important Questions

Two charges $5 \times 1 0^{- 8} C$ and $- 3 \times 1 0^{- 8} C$ are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

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Three charges $2 q, - q, - q$ are located at the vertices of an equilateral triangle. At the centre of the triangle :

Medium

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A regular hexagon of side 10 cm has a charge $5 μ C$ at each of its vertices. Calculate the potential at the centre of the hexagon.

Medium

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Four electric charges $+ q, + q, - q$ and $- q$ are placed at the corners of a square of side $2 L$ (see figure). The electric potential at point A, midway between the two charges $+ q$ and $+ q$ , is

Medium

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Two charges -q and +q are located at points (0, 0, -a) and (0, 0, a), respectively.
(a) What is the electrostatic potential at the points (0, 0, z) and (x, y, 0) ?
(b) Obtain the dependence of potential on the distance r of a point from the origin when r/a >> 1
(c) How much work is done in moving a small test charge from the point (5,0,0) to (-7,0,0) along the x-axis? Does the answer change if the path of the test charge between the same points is not along the x-axis?

Hard

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Four point charges -Q,-q,2q and 2Q are placed, one at each corner of the square. The relation between Q and q fo which the potential at the centre of the square is zero, is

Easy

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A cube of side b has a charge q at each of its vertices. Determine the potential and electric field due to this charge array at the centre of the cube.

Medium

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$27$ identical drops of mercury are charged simultaneously with the same potential of $10 V$ . Assuming the drop to the spherical, if all the charged drops are made to combine to form one large drop, then its potential will be _____ $V$ .

Easy

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Point charge $+ 4 q$ , $- q$ and $+ 4 q$ are kept on the x-axis at points $x = 0, x = a$ and $x = 4 a$ respectively. Then:

Hard

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Calculate the electric potential at the center of the square :

Medium

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Electric potential due to system of charges

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Potential Due to System of Charges

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Electric Potential due to a System of Charges

Two charges 5×10−8 C and −3×10−8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

Three charges 2q,−q,−q are located at the vertices of an equilateral triangle. At the centre of the triangle :

A regular hexagon of side 10 cm has a charge 5μC at each of its vertices. Calculate the potential at the centre of the hexagon.

Four electric charges +q,+q,−q and −q are placed at the corners of a square of side 2L (see figure). The electric potential at point A, midway between the two charges +q and +q, is

Four point charges -Q,-q,2q and 2Q are placed, one at each corner of the square. The relation between Q and q fo which the potential at the centre of the square is zero, is

A cube of side b has a charge q at each of its vertices. Determine the potential and electric field due to this charge array at the centre of the cube.

27 identical drops of mercury are charged simultaneously with the same potential of 10 V. Assuming the drop to the spherical, if all the charged drops are made to combine to form one large drop, then its potential will be _____ V.

Point charge +4q, −q and +4q are kept on the x-axis at points x=0,x=a and x=4a respectively. Then:

Calculate the electric potential at the center of the square :

Two charges $5 \times 1 0^{- 8} C$ and $- 3 \times 1 0^{- 8} C$ are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

Three charges $2 q, - q, - q$ are located at the vertices of an equilateral triangle. At the centre of the triangle :

A regular hexagon of side 10 cm has a charge $5 μ C$ at each of its vertices. Calculate the potential at the centre of the hexagon.

Four electric charges $+ q, + q, - q$ and $- q$ are placed at the corners of a square of side $2 L$ (see figure). The electric potential at point A, midway between the two charges $+ q$ and $+ q$ , is

$27$ identical drops of mercury are charged simultaneously with the same potential of $10 V$ . Assuming the drop to the spherical, if all the charged drops are made to combine to form one large drop, then its potential will be _____ $V$ .

Point charge $+ 4 q$ , $- q$ and $+ 4 q$ are kept on the x-axis at points $x = 0, x = a$ and $x = 4 a$ respectively. Then: