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>> $Electrostatic Force Physics$

Electrostatic Force Physics

Q: A polythene piece rubbed with wool is found to have a negative charge of \$3\times 10^{-7}C\$ .(a) Estimate the number of electrons transferred (from which to which?)(b) Is there a transfer of mass from wool to polythene?

Given,\$q=3\times 10^{-7}C\$\$e=1.6\times 10^{-19}C\$(a) The number of the electron,The charge is quantized, \$q=ne\$\$n=\dfrac{q}{e}=\dfrac{3\times 10^{-7}}{1.6\times 10^{-19}}\$\$n=1.875\times 10^{12}\$ electrons(b) Yes, mass of electron is transferred.We know that, \$m=9.1\times 10^{-31}kg\$The transfer of mass, \$M=nm\$\$M=1.875\times 10^{12}\times 9.1\times 10^{-31}\$\$M=17.0\times 10^{-19}kg\$

Q: A charge \$Q\$ is divided into two parts of \$q\$ and \$Q-q\$. If the Coulomb repulsion between them when they are separated, is to be maximum, the ratio of \$\dfrac{Q}{q}\$ should be:

Let parts are separated at a distance of 'd'So repulsion force,\$F=\dfrac{kq(Q-q)}{d^2}\$here, Q and d are constant so for maximum force differentiate the force w.r.t q,\$\dfrac{dF}{dq}=\dfrac{k}{d^2}{(Q-2q)}\$For maximum force, \$\dfrac{dF}{dq}=0\$ \$\dfrac{k}{d^2}(Q-2q)=0\$ \$Q-2q=0\$ \$\dfrac{Q}{q}=2\$

Q: Two free charges q and 4q are placed at a distance d apart. A third charge Q is placed between them at a distance x from charges q such that the system is in equilibrium.Then

Charges \$=q\$ and \$4q\$ distance \$=d\$third charge \$=Q\$distance \$=x\$charges \$=q\$\$\dfrac { Kq }{ { x }^{ 2 } } =\dfrac { K\left( 4q \right) }{ { \left( L-x \right) }^{ 2 } } \$or, \$\boxed { x=\dfrac { L }{ 3 } } \quad \longrightarrow \left( 1 \right) \$This is the position where the third charge should be placed. Now, we have to determine the nature and magnitude of it. For this just try to satisfy the condition of equilibrium on any one of the remaining two charges. Let's take \$+q\$ charges. The force on it due to \$+4q\$ charge is directed left wards so force on it due to the third charge should be directed right wards. This implies that the force between \$+q\$ charge and third charge should be of attractive nature. Thus the nature of the \$3rd\$ charge is \$(-ve)\$.Taking the third charge to be \$-Q\$ (say) and then on applying the condition of equilibrium on \$+q\$ charge \$\dfrac { KQ }{ { x }^{ 2 } } =\dfrac { K\left( 4q \right) }{ { L }^{ 2 } } \$\$\dfrac { KQ }{ { \left( L/3 \right) }^{ 2 } } =\dfrac { K\left( 4q \right) }{ { L }^{ 2 } } \$Putting eqn \$\Rightarrow \dfrac { 9KQ }{ { L }^{ 2 } } =\dfrac { 4Kq }{ { L }^{ 2 } } \$\$\Rightarrow 9Q=4q\$or, \$Q=\dfrac { 49 }{ q } \$.

Q: A soap bubble is given a negative charge, then its radius :

\$\bullet\$ As soap bubble is conductor, so by symmetry, the charge will distribute uniformly over the surface of bubble.\$\bullet\$ Due to mutual repulsion between like charges over the surface, they will push the parts of soap film away from each other.\$\bullet\$ Hence the radius of the bubble will Increase.Hence Correct answer is B.\$\textbf{Note:}\$ There will be a finite increase in radius of the bubble up to the point, when mutual Electrostatic repulsion of negative charges is balanced by the force of Surface tension of bubble and pressure forces from outside and inside.

Q: Equal charges \$Q\$ are placed at the four corners \$A,B,C\$ and \$D\$ of a square of length \$a\$. The magnitude of the force on the charge at \$B\$ will be

\$\begin{array}{l} force\, on\, B\, due\, to\, A={ f_{ 1 } }=\frac { { k{ Q^{ 2 } } } }{ { { a^{ 2 } } } } \\ force\, on\, C\, due\, to\, B\, ={ f_{ 2 } }=\frac { { k{ Q^{ 2 } } } }{ { { a^{ 2 } } } } \\ resul\tan t\, of\, these\, two\, forces=f=\sqrt { f_{ 1 }^{ 2 }+f_{ 2 }^{ 2 } } =\sqrt { 2 } \frac { { k{ Q^{ 2 } } } }{ { { a^{ 2 } } } } \\ force\, due\, to\, D\, ={ f_{ 3 } }=\frac { { k{ Q^{ 2 } } } }{ { 2{ a^{ 2 } } } } \\ R=net\, force=\sqrt { { f^{ 2 } }+f_{ 3 }^{ 2 }+2{ f_{ 3 } }\cos 0 } \\ =f+{ f_{ 3 } } \\ =\sqrt { 2 } \frac { { k{ Q^{ 2 } } } }{ { { a^{ 2 } } } } +\frac { { k{ Q^{ 2 } } } }{ { 2{ a^{ 2 } } } } \\ =\frac { { k{ Q^{ 2 } } } }{ { { a^{ 2 } } } } \left( { \frac { { 2\sqrt { 2 } +1 } }{ 2 } } \right) . \end{array}\$Hence, the option \$C\$ is the correct answer.

Q: A cup contains \$250g\$ of water. Find the total positive charge present in the cup of water.

Mass of water \$=250g\$Molecular mass of water \$=18g\$Number of molecules in \$18g\$ of water \$==6.02\times { 10 }^{ 23 }\$Number of moleucles in one cup of water \$=\cfrac { 250 }{ 18 } \times 6.02\times { 10 }^{ 23 }\$Each molecule fo water contains two hydrogen atoms and one oxygen atom, i.e., 10 electrons and 10 protons.\$\therefore\$ Total positive charge present in one cup of water\$=\cfrac { 250 }{ 18 } \times 6.02\times { 10 }^{ 23 }\times 10\times 1.6\times { 10 }^{ -19 }C=1.34\times { 10 }^{ 7 }C\$

Q: Five balls, numbered 1 to 5, are suspended using separate threads. Pairs (1, 2), (2, 4), (4, 1)show electrostatic attraction, while pairs (2, 3) and (4, 5) show repulsion, therefore ball 1 must be:

It should be neutral because \$2\$, and \$4\$ show attraction means have opposite charge \$2\$ too show attraction with \$1\$ means \$1\$ and \$4\$ should have same charge but as they are showing attraction too then. \$1\$ should be electricity neutral. Besides, it can be seen that ball \$2\$ is positively charged then ball \$4\$ is negatively charged because \$2\$ and \$4\$ show attraction. Now ball if ball \$2\$ is \$(+ve)\$ charged then ball \$3\$ is also \$(+ve)\$ charged because \$2\$ and \$3\$ also show repulsion and similarly ball \$5\$ is \$(-ve)\$ charged because ball \$4\$ and \$5\$ show repulsion and we left with ball \$1\$ and it never be \$(+ve)\$ by charged because ball \$2\$ and \$1\$ show attraction it means ball \$1\$ is negatively charge but it is also not possible because ball \$1\$ and \$4\$ also show attraction and in the end we know that ball \$1\$ is nor be positively or neither be negatively charge. It means be ball \$1\$ have no charge so it must be neutral.

Q: Two identical particles each of mass \$M\$ and charge \$Q\$ are placed a certain distance part. If they are in equilibrium under mutual gravitational and electric force then calculate the order of \$\dfrac{Q}{M}\$ in \$SI\$ units.

As the given system of charges are in equilibrium,hence \$F_e = F_g \$\$F_e\$ is the electric force between the charges which is repelling them while \$F_g\$ is the gravitational force of attraction.ie. \$\frac{KQ^2}{R^2} = \frac{GM^2}{R^2}\$or , \$KQ^2 = GM^2 \$ or , \$(\frac{Q}{M})^2 = (\frac{G}{K})\$or , \$\frac{Q}{M} = ({G}{K})^{\frac{1}{2}}\$as G = \$6.67 \times 10^{-11} Nm^2 / Kg^2\$and K = \$ 9 \times 10^9 Nm^2 / C^2\$\$(\frac{Q}{M})^2 = \frac{( 6.67 \times 10^{-11} )}{ ( 9 \times 10^9 )} \$or \$\frac{Q}{M} = [ 0.7411 \times 10^{-(11+9)} ] ^{\frac{1}{2}}\$ or \$\frac{Q}{M} = ( 0.7411 \times 10^{-20} ) ^{\frac{1}{2}} \$we get \$\frac{Q}{M} = 0.86 \times 10^{-10} \$ c / kg

Q: The value of the constant \$k\$ in coulomb's law is :

The Coulomb's constant, \$k=\dfrac{1}{4\pi\epsilon_0}=\dfrac{1}{4\pi\times (8.85\times 10^{-12})}=9\times 10^9\$

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Introduction to Electrostatics

Electrostatics and Electric Charge

A polythene piece rubbed with wool is found to have a negative charge of 3×10−7C . (a) Estimate the number of electrons transferred (from which to which?) (b) Is there a transfer of mass from wool to polythene?

A charge Q is divided into two parts of q and Q−q. If the Coulomb repulsion between them when they are separated, is to be maximum, the ratio of qQ​ should be:

Two free charges q and 4q are placed at a distance d apart. A third charge Q is placed between them at a distance x from charges q such that the system is in equilibrium.Then

A soap bubble is given a negative charge, then its radius :

Equal charges Q are placed at the four corners A,B,C and D of a square of length a. The magnitude of the force on the charge at B will be

A cup contains 250g of water. Find the total positive charge present in the cup of water.

Five balls, numbered 1 to 5, are suspended using separate threads. Pairs (1, 2), (2, 4), (4, 1) show electrostatic attraction, while pairs (2, 3) and (4, 5) show repulsion, therefore ball 1 must be:

A charge of 8mC is located at the origin. Calculate the work done in taking a small charge of −2×10−9C from a point P(0,0,3cm) to a point Q(0,4cm,0), via a point R(0,6cm,9cm).

Two identical particles each of mass M and charge Q are placed a certain distance part. If they are in equilibrium under mutual gravitational and electric force then calculate the order of MQ​ in SI units.

The value of the constant k in coulomb's law is :

A polythene piece rubbed with wool is found to have a negative charge of $3 \times 1 0^{- 7} C$ .
(a) Estimate the number of electrons transferred (from which to which?)
(b) Is there a transfer of mass from wool to polythene?

A charge $Q$ is divided into two parts of $q$ and $Q - q$ . If the Coulomb repulsion between them when they are separated, is to be maximum, the ratio of $\frac{Q}{q}$ should be:

Equal charges $Q$ are placed at the four corners $A, B, C$ and $D$ of a square of length $a$ . The magnitude of the force on the charge at $B$ will be

A cup contains $250 g$ of water. Find the total positive charge present in the cup of water.

Five balls, numbered 1 to 5, are suspended using separate threads. Pairs (1, 2), (2, 4), (4, 1)
show electrostatic attraction, while pairs (2, 3) and (4, 5) show repulsion, therefore ball 1 must be:

A charge of $8 m C$ is located at the origin. Calculate
the work done in taking a small charge of $- 2 \times 1 0^{- 9} C$ from a point
$P (0, 0, 3 c m)$ to a point $Q (0, 4 c m, 0)$ , via a point $R (0, 6 c m, 9 c m)$ .

Two identical particles each of mass $M$ and charge $Q$ are placed a certain distance part. If they are in equilibrium under mutual gravitational and electric force then calculate the order of $\frac{Q}{M}$ in $S I$ units.

The value of the constant $k$ in coulomb's law is :