The electrostatic force on a small sphere of charge \$0.4\mu C\$ due to another small sphere of charge \$-0.8\mu C\$ in air is 0.2 N.(a) What is the distance between the two spheres? (b) What is the force on the second sphere due to the first?

(a) Electrostatic force on the first sphere, \$F = 0.2 N\$Charge on this sphere, \$q_1=0.4\mu C = 0.4\times 10^{-6}C\$Charge on the second sphere, \$q_2=-0.8\mu C=-0.8\times 10^{-6}C\$Electrostatic force between the spheres is given by the relation,\$\displaystyle F=\frac{q_1q_2}{4\pi \varepsilon_0r^2}\$And, \$\displaystyle \frac{1}{4\pi \varepsilon_0} = 9\times 10^9N\,m^2C^{-2}\$Where, \$\varepsilon_0=\$ Permittivity of free space\$r^2=\displaystyle \frac{q_1q_2}{4\pi \varepsilon_0F}\$ \$\displaystyle =\frac{04\times 10^{-6}\times 8\times 10^{-6}\times 9\times 10^{9}}{0.2}\$ \$=144\times 10^{-4}\$ \$r =\sqrt{144\times 10^{-4}}\$ \$=0.12m\$The distance between the two spheres is 0.12 m.(b) Equal and Opposite Force acts on the other sphere (By newton's Third Law). Hence 0.2 N attractive.

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Coulomb's Law - Problem L1

Q: What is the force between two small charged spheres having charges of \$2\times 10^{-7}\$C and \$3\times 10^{-7}\$C placed 30 cm apart in air?

Charge on the first sphere, \$ q_1= 2 \times 10^{-7}\$ CCharge on the second sphere, \$ q_2= 3 \times 10^{-7}\$ CDistance between the spheres, \$r = 30 cm = 0.3 m\$\$F=\displaystyle \frac{q_1q_2}{4\pi \varepsilon_0r^2}\$Where, \$\varepsilon_0=\$ Permittivity of free space\$\displaystyle \frac{1}{4\pi\varepsilon_0}=9\times 10^9\, N\,m^2C^{-2}\$\$\displaystyle F=\frac{9\times 10^9\times 2\times 10^{-7}\times 3\times 10^{-7}}{(0.3)^2}=6\times 10^{-3}N\$Hence, force between the two small charged spheres is \$ 6 \times 10^{-3}\$ N. The charges are of same nature. Hence, force between them will be repulsive.

Q: Two point charges A and B, having charges +Q and -Q respectively, are placed at certain distance apart and force acting between them is F. If 25% charge of A is transferred to B, then force between the charges becomes

\$F=\dfrac{KQ^2}{r^2}\$If 25% of charge of A transferred to B then \$q_A = Q - \dfrac{Q}{4} = \dfrac{3 Q}{4} \$ and \$q_B = -Q + \dfrac{Q}{4} = \dfrac{-3Q}{4}\$\$F_1 = \dfrac{kq_A q_B}{r^2}\$\$F_1 = \dfrac{k \left(\dfrac{3Q}{4} \right)^2}{r^2}\$\$F_1 = \dfrac{9}{16} \dfrac{kQ}{r^2}\$\$\implies F_1 = \dfrac{9F}{16}\$

Q: The electrostatic force of repulsion between two equal positively charged ions is \$3.7\times 10^{-9}N\$, when they are separated by a distance of \$5\overset {\circ}{A}\$. How many electrons are missing from each ion?

Given, the force of repulsion between two equally positively charged ions \$=3.7\times 10^{-19}\, N\$Distance of separation, \$d=5\$Å \$=5\times 10^{10}\, m\$To find the number of electrons missing from each ion.Let us suppose charge is \$q_{1}=q_{2}=q\$Let \$n\$ be the number of electrons missing.Using Coulomb's Law, \$F=\dfrac{1}{4\pi \varepsilon _{0}}\dfrac{q_{1}q_{2}}{d^{2}}\$\$F=\dfrac{1}{4\pi \varepsilon _{0}}\dfrac{q^{2}}{d^{2}}\$So, \$q^{2}=4\pi \varepsilon _{0}\times F\times d^{2}\$\$=\dfrac{1}{9\times 10^{9}}\times 3.7\times 10^{-9}\times (5\times 10^{-10})^{2}\$\$=10.28\times 10^{-38}\$\$q=3.2\times 10^{-19}\, C\$As, \$q=ne\$So, \$n=\dfrac qe=\dfrac{3.2\times 10^{-19}}{1.6\times 10^{-19}}\$\$n=2\$

Q: Suppose the charge of a proton and an electron differ slightly. One of them is -e, the other is \$(e+\Delta e)\$. If the net of electrostatic force and gravitational force between two hydrogen atoms placed at a distance d (much greater than atomic size) apart is zero, then \$\Delta e\$ is of the order of [Given mass of hydrogen \$m_h=1.67\times 10^{-27}Kg\$]

Given : \$m_h= 1.67\times 10^{-27} \ kg\$Net charge on hydrogen atom \$q = (e+\Delta e)-e = \Delta e\$Let the distance between them be \$d\$.Equating gravitational force and the electrostatic force, we get\$\dfrac{Gm_h^2}{d^2} = \dfrac{k (\Delta e)^2}{d^2}\$Or \$\dfrac{(6.67\times 10^{-11})(1.67\times 10^{-27})^2}{d^2} = \dfrac{(9\times 10^9)(\Delta e)^2}{d^2}\$\$\implies \ \Delta e = 2.06\times 10^{-37} \ C\$The correct answer is option D.

Q: (a) Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is \$6.5\times 10^{-7}\$ C? The radii of A and B are negligible compared to the distance of separation.(b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?

(a) Charge on sphere A, \$q_A=6.5 \times 10^{-7} C\$ Charge on sphere B, \$q_B=6.5\times 10^{-7} C\$Distance between the spheres, \$r = 50 cm = 0.5 m\$Force of repulsion between the two spheres,\$F=\displaystyle \frac{q_Aq_B}{4\pi\in_0r^2}\$Where, \$\in_0=\$ Free space permittivity\$\displaystyle \frac{1}{4\pi\in_0} = 9\times 10^9N\,m^2C^2\$\$\therefore \displaystyle F = \frac{9\times 10^9\times (6.5\times 10^{-7})^2}{(0.5)^2}\$ \$=1.52\times 10^{-2}N\$Therefore, the force between the two spheres is \$1.52\times 10^{-2}N\$(b) After doubling the charge, Charge on sphere A, \$q_A=2 \times 6.5 \times 10^{-7}C=1.3 \times 10^{-6}C\$ Charge on sphere B, \$q_B=2\times 6.5\times 10^{-7}C = 1.3\times 10^{-6}C\$Now, if the distance between the sphere is halved, then\$r=\frac{0.5}{2}=0.25m\$Force of repulsion between the two sphere,\$\displaystyle F=\frac{q_Aq_B}{4\pi\in_0r^2}=\frac{9\times 10^9\times 1.3\times 10^{-6}\times 1.3\times 10^{-6}}{(0.25)^2}=0.24336\,N\$Therefore, the force between the two sphere is approximately 0.243 N.

Q: Charge \$Q\$, is divided into two parts which are then kept some distance apart. The force between them will be maximum if the two parts are having the charge.

\$\textbf{Step 1: Calculation of force }\$ Let one part of the charge be \$q\$. Thus other part of the charge is \$Q-q\$From Coulomb's law:Electrostatic force between them \$F = \dfrac{kq(Q-q)}{r^2}\$\$\textbf{Step 2: Condition of maximum force }\$ For maximum force, \$\dfrac{dF}{dq} = 0\$\$\therefore\$ \$\dfrac{d}{dq}\bigg(\dfrac{kq(Q-q)}{r^2}\bigg)= 0\$Or, \$\dfrac{d}{dq}(qQ-q^2) = 0\$Or, \$Q-2q = 0\$ \$\Rightarrow\$ \$q = \dfrac{Q}{2}\$Hence force between them is maximum if the two parts have equal charge \$Q/2\$ each. So, Option A is correct

Q: Two similarly and equally charged identical metal spheres A and B repel each other with a force of \$2 \times 10^{-5} N\$. A third identical uncharged sphere C is touched with A and then placed at the midpoint between A and B. Find the net electric force on C :

Let the charge on each sphere be q then the force on each sphere is \$F=\dfrac{kqq}{r^2}=\dfrac{kq^2}{r^2}=2\times 10^{-5} .....(1)\$ When sphere C is touched by sphere A, charge q gets equally divided between the two.Thus, \$F_{CA}=\dfrac{k(q/2)(q/2)}{r^2/4}=\dfrac{kq^2}{r^2}\$and \$F_{CB}=\dfrac{k(q/2)(q)}{r^2/4}=\dfrac{2kq^2}{r^2}\$Net force \$=F_{CB}-F_{CA}=\dfrac{2kq^2}{r^2}-\dfrac{kq^2}{r^2}=\dfrac{kq^2}{r^2}=2\times 10^{-5} N \$ (using (1))As \$F_{CB}>F_{CA}\$ so net force is away sphere B.

Q: The force between two charges 0.06 m apart is \$5\ N\$. If each charge is moved towards the other by \$0.01\ m\$, then the force between them will become

\$\textbf{Step 1: Initial Force between two charges }\$Using Coulomb's law, \$F = \dfrac {Kq_{1}q_{2}}{r^{2}}\$ \$\Rightarrow 5 = \dfrac {Kq_{1}q_{2}}{(0.06)^{2}} \$ \$.... (1)\$\$\textbf{Step 2: New Force between the charges}\$ When each charge is moved towards the other by \$0.01m\$ , the new distance between them is \$r'= 0.06m-2\times 0.01m= 0.04m\$ \$F' = \dfrac {Kq_{1}q_{2}}{(0.04)^{2}}\$ \$.....(2)\$ \$\textbf{Step 3: Solving Equations}\$Dividing \$(2)\$ by \$(1)\Rightarrow\$ \$\dfrac{F'}{5} = \dfrac {(0.06)^2}{(0.04)^{2}}\$ \$\Rightarrow F' = 11.25\ N\$The new force would be \$11.25\ N\$Hence, Option(B) is correct.

Q: When the distance between two charged particles is halved, then the force between them becomes :

Earlier force between two particles is \$F= \dfrac{kq_1q_2}{r^2}\$ When distance is halved force becomes \$F'=\dfrac{kq_1q_2}{(r/2)^2}=4 \dfrac{kq_1q_2}{r^2}=4 F\$

Physics

example

Force between two point charges

A charge

q_{1}

exerts a force of 45N on a test charge

q_{2} = 1 0^{- 5} C

located at a point 0.2m from

q_{1}

. The magnitude of

q_{1}

is:
We know

F = \frac{k q _{1} q _{2}}{r ^{2}}

\Rightarrow 45 = \frac{( 9 \times 1 0 ^{9} ) ( q _{1} ) ( 1 0 ^{- 5} )}{( 0 \cdot 2 ) ^{2}}

\Rightarrow q_{1} = \frac{4 5 \times 0 \cdot 2 \times 0 \cdot 2}{9 \times 1 0 ^{9} \times 1 0 ^{- 5}}

\Rightarrow q_{1} = 2 \times 1 0^{- 5} C

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Problem on Coulomb's Law Involving SHM

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