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>> $Electric Field due to System of Charges$

Electric Field due to System of Charges

Q: Four charges each equal to \$Q\$ are placed at the four corners of a square and a charge \$q\$ is placed at the centre of the square. If the system is in equilibrium, the value of \$q\$ is :

\$ \textbf {Step 1 : Calculate all the required distances between the charges :} \$Let the side length of the square be \$a\$.Refer Figure, In \$\triangle ACD\$, \$\Rightarrow AD^2 +CD^2 = AC^2\$\$\Rightarrow a^2+ a^2 = AC^2\$ \$\Rightarrow AC = \sqrt{2}a\$However, \$ OC =\dfrac{AC}{2} \$ \$=\cfrac{a}{\sqrt{2}}\$\$ \textbf {Step 2 : Equilibrium of central charge} \$By symmetry, Force on central charge will be equal and opposite due to the diagonally opposite charges, which will cancel each other.Hence, Net force on central charge will always be zero, irrespective of value of charge \$q\$.Therefore, to find value of \$q\$ we have to check equilibrium of any one charge at the corner.\$ \textbf {Step 3 : Force due to all the charges at point C } \$Force on charge at \$C\$ due to \$B\$, \$F_1 = \dfrac{K Q^2}{a^2}\$ Force on charge at \$C\$ due to \$D\$, \$F_2 = \dfrac{K Q^2}{a^2}\$ Force on charge at \$C\$ due to \$A\$, \$F_4 = \dfrac{K Q^2}{AC^2} = \dfrac{KQ^2}{2a^2}\$ Force on charge at \$C\$ due to \$q\$ at centre, \$F_3 = \dfrac{K qQ}{OC^2} = \dfrac{2KqQ}{a^2}\$ \$ \textbf {Step 4 : Apply the equilibrium condition at C :} \$For the system to be in equilibrium, net force acting on charge at C must be zero.So, \$\vec{F_3} + \vec{F_1} + \vec{F_2} + \vec{F_4} =0\$ Resultant of \$F_1\$ and \$F_2\$ (Along OC, by symmetry) = \$\sqrt{F_1^2 +F_2^2} = \sqrt{2} F_1\$ Since \$( |F_1| = |F_2|)\$Also, \$F_3\$ & \$F_4\$ are along OC, Therefore, magnitudes of sum of these forces should be zero. \$\Rightarrow \$ \$|F_3| + \sqrt{2} |F_1| + |F_4| =0\$ \$\Rightarrow \dfrac{2K qQ}{a^2}\$ \$+ \sqrt{2} \dfrac{K Q^2}{a^2}\$ \$+ \dfrac{K Q^2}{2a^2}=0\$ \$\Rightarrow 2q = -(\sqrt{2}Q + \dfrac{Q}{2})\$ \$\Rightarrow q = -\dfrac{Q}{4} (1 + 2\sqrt{2})\$

Q: A charge Q is placed at the centre of the line joining two equal charges q. Show that the system of three charges will be in equilibrium if \$Q=\dfrac{-q}{4}\$.

\$\textbf{Step 1: Equilibrium of central charge Q }\$\$\textbf{[Ref. Fig. 1]}\$For the system to be in equilibrium net force on each charge should be zero.By Symmetry, both outer charges \$q\$ will exert equal and opposite force on \$Q\$, which will cancel out. As shown in Figure 1.Hence Net force on \$Q\$ will always be zero irrespective of value of charge.\$\textbf{Step 2: Equilibrium of outer charge q }\$\$\textbf{[Ref. Fig. 2]}\$Net force on charge \$q\$ at the right, due to other two charges should be zero. \$F_{q} = \dfrac{KQq}{x^2} + \dfrac{Kq^2}{(2x)^2} = 0\$ \$\Rightarrow\ \ \dfrac{KQq}{x^2} = -\dfrac{Kq^2}{4x^2}\$ \$\Rightarrow\ \ Q= \dfrac{-q}{4}\$

Q: Three charges +4q, Q and q are placed in a straight line of length \$l\$ at points at distances 0, \$l/2\$, and \$l\$ respectively. What should be Q in order to make the net force on q to be zero?

Answer is A.The force exerted on the charge q,Q is given as follows.\$F(q,Q)=k[\frac { qQ }{ l^{ 2 }/4 } ]=\frac { 4kqQ }{ l^{ 2 } } \$.The force exerted on the charge q, 4q is given as follows.\$F(q,4q)=k[\frac { 4q^{ 2 } }{ l^{ 2 } } ]=\frac { 4kq^{ 2 } }{ l^{ 2 } } \$.The net force = F(q,Q) + F(q,4q) = 0 \$\frac { 4kqQ }{ l^{ 2 } } +\frac { 4kq^{ 2 } }{ l^{ 2 } } = 0\$So, \$qQ+q^{ 2 }=0\$.q[Q + q] =0 As, q cannot be = 0 Q + q =0 Therefore, Q = - q Hence, Q should be -q in order to make the net force on q to be zero.

Q: Four point charges \$q_A = 2\mu C, q_B = -5 \mu C, q_C = 2 \mu C\$, and \$q_D = -5 \mu C \$are located at the corners of a square \$ABCD\$ of side \$10\ cm\$. What is the force on a charge of \$1 \mu C\$ placed at the centre of the square?

The given figure shows a square of side 10 cm with four charges placed at its corners. O is the centre of the square.Where,(Sides) \$AB = BC = CD = AD = 10 cm\$(Diagonals) \$AC = BD =10\sqrt{2}\,cm\$ \$AO = OC = DO = OB =5\sqrt{2}\,cm\$A charge of amount \$1\mu C\$ is placed at point O.Force of repulsion between charges placed at corner A and centre O is equal in magnitude but opposite in direction relative to the force of repulsion between the charges placed at corner C and centre O. Hence, they will cancel each other. Similarly, force of attraction between charges placed at corner B and centre O is equal in magnitude but opposite in direction relative to the force of attraction between the charges placed at corner D and centre O. Hence, they will also cancel each other. Therefore, net force caused by the four charges placed at the corner of the square on \$1\mu C\$ charge at centre O is zero.

Q: A charge Q is placed at each of the two opposite corners of a square. A charge q is placed at each of the other two comers. If the resultant force on each charge q is zero, then :

The net force on q at one corner is zero if \$\vec{F_1}+\vec{F_2}+\vec{F_3}=0\$or \$F_1\cos45^o\hat i-F_1\sin45^o\hat j-F_2\hat j+F_3\hat i=0\$ so, \$F_1\cos45^o=-F_3 ....(1)\$ and \$F_1\sin45^o=-F_2 ....(2)\$using (1) , \$\dfrac{kq^2}{(\sqrt{2}a)^2}\times \dfrac{1}{\sqrt 2}=-\dfrac{kqQ}{a^2}\$or \$q=-2\sqrt 2 Q\$

Q: An electron falls through a distance of \$1.5 cm\$ in a uniform electric field of magnitude \$2 \times {10^4}N/C.\$ Now the direction of the field is reversed keeping the magnitude and unchanged and a proton falls through the same distance. Compute the time of fall in each case, neglecting gravity.

ATQ, \$S=1.5cm=1.5\times { 10 }^{ -2 }m\$ \$u=0m/s\$\${ m }_{ e }\$ : mass of electron \$=9.10\times { 10 }^{ -31 }Kg\$\${ m }_{ p }\$ : mass of proton \$=1.672\times { 10 }^{ -27 }Kg\$from coloumb's force, \$\vec { F } =q\vec { E } \$and given \$\vec { E } =2\times { 10 }^{ 4 }N/c\$force proton \${ \vec { F } }_{ (+) }=1.6\times { 10 }^{ -19 }\times 2\times { 10 }^{ 4 }=3.2\times { 10 }^{ -19 }N\$force on electron, \${ \vec { F } }_{ (-) }=-1.6\times { 10 }^{ -19 }\times 2\times { 10 }^{ 4 }=-3.2\times { 10 }^{ -15 }N\$From Newton's Second law, \$\vec { F } =m\vec { a } \$\$\therefore \$ for proton, \${ a }_{ \left( + \right) }=\dfrac { { \vec { F } }_{ (+) } }{ { m }_{ p } } =1.913\times { 10 }^{ 12 }m/{ s }^{ 2 }\$for electron, \${ a }_{ \left( - \right) }=\dfrac { { \vec { F } }_{ \left( - \right) } }{ { m }_{ e } } =-0.35\times { 10 }^{ 16 }m/{ s }^{ 2 }\$from equ of motion, \$S=ut+\dfrac { 1 }{ 2 } { at }^{ 2 }\quad \longrightarrow (1)\$ATQ, The direction of the field was reversed, hence we will ignore the sign convention for acceleration.\$\therefore \$ Substituting values in equ (1) for proton,\$1.5\times { 10 }^{ -2 }=\dfrac { 1 }{ 2 } \times 1.913\times { 10 }^{ 12 }\times { t }^{ 2 }\$\$1.568\times { 10 }^{ -14 }={ t }^{ 2 }\$\$\Rightarrow \quad t=1.252\times { 10 }^{ -7 }seconds\$Substituting values in equ(1) for electron,\$1.5\times { 10 }^{ -2 }=\dfrac { 1 }{ 2 } \times 3.5\times { 10 }^{ 15 }\times { t }^{ 2 }\$\$0.857\times { 10 }^{ -17 }={ t }^{ 2 }\$\$\Rightarrow \quad { t }_{ 0 }=2.927\times { 10 }^{ -19 }seconds\$

Q: Plot a graph showing the variation of coulomb force \$(F)\$ versus \$\dfrac{1}{r^2}\$ where \$r\$ is the distance between the two charges of each pair of charge \$(1 \ \mu C, 2 \ \mu C)\$ and \$(1 \ \mu C, -3 \ \mu C)\$. Interpret to that graphs obtained.

\$\textbf{Step 1: Calculation of force for first pair of charges}\$ \$1μC\$ and \$2μC\$According to Coulomb's law:\$F=K \dfrac{q_1q_2}{r^2}\$ \$=K \dfrac{1\mu C\times 2\mu C}{r^2}\$\$\Rightarrow\$ \$F=2K({\dfrac{1}{r^2}})\$To draw graph of \$F\$ versus \$1/r^2\$On comparing above equation with \$y=mx\$ where \$m\$ is the slopeWe get, Slope of equation \$m_A=2K\$ \$....(1)\$Both charges will repel each other, as they are of same sign, hence slope is positive.\$\textbf{Step 2: Calculation of force for second pair}\$ \$2μC\$ and \$-3μC\$\$F'=K \dfrac{q'_1q'_2}{r^2}\$ \$=K \dfrac{1\mu C\times -3\mu C}{r^2}\$ \$\Rightarrow F'=-3K(\dfrac{1}{r^2})\$To draw graph of \$F'\$ versus \$1/r^2\$On comparing above equation with \$y=mx\$, where \$m\$ is the slopeWe get, Slope of equation \$m_B=-3K\$ \$....(2)\$Both charges are of opposite sign, therefore they will attract each other, that's why slope is negative. \$\textbf{Step 3: Comparing slopes and drawing the graph }\$ \$y=mx\$ is the equation of straight lineFrom equation \$(1)\$ and \$(2)\$, we observe thatMagnitude of slope of B is greater than the magnitude of slope of A i.e. \$m_B > m_A\$Hence, graph between F and\$ \dfrac{1}{r^2}\$ will be straight line where line \$B\$ has higher slope than line \$A\$ as shown in figure above.

Q: Four point charges \$Q, q, Q\$ and \$q\$ are placed at the corners of a square of side \$'a'\$ as shown in the figure.Find the(a) resultant electric force on a charge \$Q\$, and(b) potential energy of this system.

(a)Electric force=\$\dfrac{k q_{1} q_{2}}{r^{2}}\$Electric force on Q by other charge Q=\$\dfrac{kQ^{2}}{2a^{2}}\$force on Q due to both q=\$\sqrt{2}\dfrac{kqQ}{a^{2}}\$The resultant electric force =\$\dfrac{kQ^{2}}{2a^{2}}\$+\$\sqrt{2}\dfrac{kqQ}{a^{2}}\$=\$\dfrac{kQ}{2a^{2}}(Q+2\sqrt{2}q)\$(b)Potential energy of two charges =\$\dfrac{k q_{1} q_{2}}{r}\$potential energy of system=\$4\dfrac{k Qq}{a}\$+\$\dfrac{kQ^{2}}{\sqrt{2}a}\$+\$\dfrac{kq^{2}}{\sqrt{2}a}\$

Q: Two charges \$q\$ and \$-3q\$ are fixed on x-axis separated by distance \$d\$. Where should a third charge \$2q\$ be placed from \$A\$ such that it will not experience any force?

Let a charge \$2q\$ be placed at \$P\$, at a distance \$l\$ from \$A\$ where charge \$q\$ is placed, as shown in figureThe charge \$2q\$ will not experience any force, when force of repulsion on it due to \$q\$ is balanced by force of attraction on it due to \$-3q\$ at \$B\$ where \$AB=d\$or \$\cfrac { \left( 2q \right) \left( q \right) }{ 4\pi { \varepsilon }_{ 0 }{ l }^{ 2 } } =\cfrac { \left( 2q \right) \left( 3q \right) }{ 4\pi { \varepsilon }_{ 0 }{ (l+d) }^{ 2 } } \$\${ (l+d) }^{ 2 }=3{ l }^{ 2 }\quad or\quad 2{ l }^{ 2 }-2ld-{ d }^{ 2 }=0\quad \$\$\therefore l=\cfrac { 2d\pm \sqrt { 4{ d }^{ 2 }+8{ d }^{ 2 } } }{ 4 } =\cfrac { d }{ 2 } \pm \cfrac { \sqrt { 3 } d }{ 2 } \Rightarrow l=\cfrac { d+\sqrt { 3 } d }{ 2 } \$If charge \$2q\$ is placed between \$A\$ and \$B\$ then \$l=\cfrac { (\sqrt { 3 } -1) }{ 2 } d\$

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