An electron moves with a velocity \vec{v} in an electric field \vec{E} . If the angle between \vec{v} and \vec{E} is neither 0 nor \pi , then the path followed by the electron is a(an):

The perpendicular component of velocity is given by: v_y=v \sin\theta Now in the X direction, the electron experiences a force of E(-e) So, F= -eE or, a =\dfrac {-Ee}{m} \therefore velocity in X direction: v_x= u + at = v cos\theta-\dfrac {Ee}{m}\times t We know, y=\int v_ydt and x=\int (v_x)dt \Rightarrow y=v \sin\theta \cdot t, x=v cos\theta \cdot t-\dfrac {Ee}{m}\dfrac {t^2}{2} Substituting t=\dfrac {y}{v sin\theta} , we get x=v \cos\theta \dfrac { y}{v sin\theta}-\dfrac {Ee}{m_2}\cdot \dfrac {y^2}{v^2 sin^2\theta} \Rightarrow x=y \cot\theta -\dfrac {Ee}{2m v^2sin^2\theta}y^2 , where \theta is the initial angle of v with E . \therefore path followed is a parabola

Two non-conducting spheres of radii \mathrm{R}_{1} and \mathrm{R}_{2} carrying uniform volume charge densities +\rho and -\rho , respectively, are placed such that they partially overlap, as shown in the figure. At all points in the overlapping region:

Let the vector joining the centers of the two spheres be \vec{d} and consider a point P in the overlap region, whose position vector be \vec{r_1} and \vec{r_2} with respect to the centers of each sphere respectively.Thus, by triangle law of vector addition, \vec{r_1}-\vec{r_2} = \vec{d} Electric field at P due to the positively charged sphere is E_1 = \displaystyle \frac{1}{4\pi\epsilon_0}\frac{q}{r_1^3}\vec{r_1} Here, q is the total charge enclosed in the sphere of radius r_1 Thus, q = \displaystyle \rho \times \frac{4}{3}\pi r_1^3 Hence, E_1 = \displaystyle \frac{\rho}{3\epsilon_0}\vec{r_1} Similarly, Electric field due to negatively charged sphere is E_2 = \displaystyle -\frac{\rho}{3\epsilon_0}\vec{r_2} Thus, net electric field at P is E = E_1+E_2 = \displaystyle \frac{\rho}{3\epsilon_0}(\vec{r_1}-\vec{r_2}) = \frac{\rho}{3\epsilon_0}\vec{d} Both, direction and magnitude of the electric field are constant in the overlap region.

A circular ring of radius R with uniform positive charge density \lambda per unit length is fixed in the Y-Z plane with its centre at the origin O . A particle of mass m and positive charge q is projected from the point P\left( \sqrt { 3 } R,0,0 \right) on the positive X-axis directly towards O , with initial velocity v . The smallest value of the speed v such that the particle does not return to P is \sqrt { \cfrac { \lambda q }{ x{ \varepsilon }_{ 0 }m } } . Find x .

Ring and Charge particle has a positive charge so both will repel each other.But when Velocity of projection of charge q is given in such a way that it just cross the center of the ring then it will never come back because of repulsion force.Potential at distance r from circumference of curve is V=\dfrac{k\lambda2\pi R}{r} Initially, r^2=(\sqrt 3R)^2+R^2\Rightarrow r=2R finally r=R (particle at center)From conservation of energy:Loss in KE = Gain in Electrical potential energy.so, \dfrac{m(v_1^2-v_2^2)}{2}=\dfrac{k\lambda2\pi R}{R}-\dfrac{k\lambda2\pi R}{2R} \dfrac{m(v_1^2-0)}{2}=\dfrac{k\lambda2\pi }{1}-\dfrac{k\lambda2\pi }{2} \dfrac{mv_1^2}{2}=\dfrac{k\lambda2\pi }{2} K=\dfrac1{4 \pi \epsilon_0} \Rightarrow v=\sqrt{\dfrac{\lambda q}{2\epsilon_o m}} On comparing from question x=2

A particle of mass m and negative charge q is thrown in a gravity free space with speed u from the point A in the large nonconducting charged sheet with surface charge density \sigma , as shown in figure. The maximum distance from A on sheet where the particle can strike is \cfrac { x{ \epsilon}_{ 0 }{ u }^{ 2 }m }{ q\sigma } . Find x .

Electric potential due to large non conducting sheet = \dfrac {\sigma} {2 \epsilon_{o}} Force on charge = qE = \dfrac {\sigma q} {2 \epsilon_{o}} towards sheetacceleration, a = F/m = \dfrac {\sigma q} {2m \epsilon_{o}} initial velocity perpendicular to sheet = u \sin\theta velocity parallel to sheet = u \cos \theta Time of flight = \dfrac {2 u \sin \theta}{a} Distance traveled parallel to sheet = \dfrac {2 u \sin \theta u \cos \theta }{a} = \dfrac {u^{2} \sin (2\theta)}{a} So distance travelled will be maximum for \theta =45^\circ and equal to \dfrac{u^{2} 2 m \epsilon_{o}}{\sigma q} = \dfrac{2u^{2} m \epsilon_{o}}{\sigma q} therefore x = 2

The flux of \overrightarrow{g} through this surface is given by :

From the inverse square dependence of forces, we can have an analogy of electric and gravitational forces.The charge is analogous to mass.Gravitation states F\propto -m_1m_2 and in Electrostatics, F\propto q_1q_2 The proportionality constant is G in gravitation and \displaystyle \frac{1}{4\pi\epsilon_0} in electrostatics.Thus, G\sim\displaystyle \frac{1}{4\pi\epsilon_0} and m\sim q Gauss' law for electrostatics is \displaystyle\oint \vec{E}.\vec{ds} = \frac{q_\textrm{encl}}{\epsilon_0} = 4\pi \frac{q_\textrm{encl}}{4\pi\epsilon_0} The analogous quantity for electric field is \vec{g} Thus, the Gauss' law for gravitation is \displaystyle \oint\vec{g}.\vec{ds} = -4\pi Gm_\textrm{encl} But, the mass enclosed in the surface is m Thus, \displaystyle \oint \vec{g}.\vec{dS} =-4\pi Gm

Electric Charges And Fields - Types Of Questions

Q: A charge Q is divided into two parts q_{1} and q_{2} such that they experience maximum force of repulsion when separated by certain distance. The ratio of Q, q_{1} and q_{2} is :

\textbf{Step 1: Divide Q} Q is divided into two parts q_1 and q_2 \therefore Q = q_{1} + q_{2} ....(1) \textbf{Step 2: Force of repulsion between}\ q_{1}\ \textbf{and}\ q_{2} Let r be the distance between charges. Using Coulomb's law, F = \dfrac{k q_{1} q_{2}}{r^2} = \dfrac{k q_{1} (Q - q_{1})}{r^2} ....(2) \textbf{Step 3: Condition for force to be maximum} The repulsive force between the charges will be maximum if, \dfrac{dF}{dq_{1}} = 0 \Rightarrow\dfrac{k}{r^2} (Q - 2q_{1}) = 0 \Rightarrow q_{1} = \dfrac{Q}{2} Equation (1)\Rightarrow q_{2} = Q - \dfrac{Q}{2} = \dfrac{Q}{2} \textbf{Step 4: Ratio of}\ Q : q_{1} : q_{2} Q : q_{1} : q_{2} = Q : \dfrac{Q}{2} : \dfrac{Q}{2} = 1 : \dfrac{1}{2} : \dfrac{1}{2} \Rightarrow Q : q_{1} : q_{2} = 2 : 1 : 1 Hence, Option (D) is correct.

Q: A spherical metal shell A of radius R_{A} and a solid metal sphere B of radius R_{B}(< R_{A}) are kept far apart and each is given ' +Q '. Now they are connected by thin metal wire. Then:

The potential for both the objecs is same so, \displaystyle \frac{KQ_{A}}{R_{A}} = \dfrac{KQ_{B}}{R_{B}} or \dfrac{Q_A}{Q_B}=\dfrac{R_A}{R_B} As R_BQ_B Above equation represents B is correct option.From Gauss law the electric field inside a spherical shell is zero so option A is correct.Now \sigma_A=\dfrac{Q_A}{4\pi R_A^2} and \sigma_B=\dfrac{Q_B}{4\pi R_B^2} Thus \dfrac{\sigma_A}{\sigma_B}=\dfrac{Q_A}{Q_B}\times(\dfrac{R_B}{R_A})^2=\dfrac{R_A}{R_B}\times(\dfrac{R_B}{R_A})^2=\dfrac{R_B}{R_A} Thus option C is also correct.Electric fields on the surface of shell and sphere are E_A=\dfrac{\sigma_A}{\epsilon_o} and E_B=\dfrac{\sigma_B}{\epsilon_o} Thus \dfrac{E_A}{E_B}=\dfrac{\sigma_A}{\sigma_B}<1 or E_A<E_B So option D is also correct.

Q: A particle of mass m and charge q is fastened to one end of a light string of length l . The other end of the string is fixed to the point O . The whole system lies on a frictionless horizontal plane. Initially, the mass is at rest at A . A uniform electric field in the direction shown is switched on. Then,

Work done by tension = 0Displacement = l (1 - cos 60^0) Work done by Electric field = qE \times l (1 - cos 60^0) (B) \sqrt {qEl}{2} = \frac {1}{2}mV^2 \Rightarrow V = \sqrt {\frac {qEl}{m}} (C) T - qE = \frac {mV^2}{l} \Rightarrow T = 2qE

Q: A point charge +q & mass 100 gm experiences a force of 100N at a point at a distance 20cm from a long infinite uniformly charged wire. If it is released its speed is 20\sqrt { l{ n }x } m/s when it is at a distance 40cm from wire. Find x .

For the inifinite uniform charged wire, E=\lambda/2\pi\epsilon_o r at r=20cm , Force F=\lambda q/2\pi\epsilon_o\times 0.2=100N acceleration, a=\lambda q/2\pi\epsilon_o\times0.1=vdv/dr so, v^2/2=\dfrac{\lambda}{2\pi\epsilon_o\times0.1}\int_{0.2}^{0.4}lnr=200ln2 \Rightarrow v=20\sqrt{ln 2} Thus, x=2

Q: The length of each side of a cubical closed surface is l . If charge q is situated on one of the vertices of the cube, then if the flux passing through shaded face of the cube is \cfrac { q }{ x{ \in }_{ 0 } } . Find x :

The cube has six surfaces . When a charge q at center of the cube , the flux through each surface is \displaystyle \phi=\frac{q}{6\epsilon_0} When q is placed at one of the vertices of the cube, the charge inside the cube is q/4 .now flux \displaystyle \phi=\frac{(q/4)}{6\epsilon_0}=\frac{q}{24\epsilon_0} thus, x=24

Q: A particle of mass m and charge -q moves along a diameter of a uniformly charged sphere of radius R and carrying a total charge +Q . The frequency of S.H.M of the particle if the amplitude does not exceed R , is \cfrac { 1 }{ 2\pi } \sqrt { \cfrac { qQ }{ x\pi { \varepsilon }_{ 0 }m{ R }^{ 3 } } } . Find x .

Using Gauss's law electric field inside solid sphere , \displaystyle E.4\pi r^2=\frac{\rho.(4/3)\pi r^3}{\epsilon_0} where \rho= charge density and here Q=\rho(4/3)\pi R^3 .Thus, \displaystyle E=\frac{Qr}{4\pi\epsilon_0 R^3} The force on charge -q is \displaystyle F=-qE=-\frac{qQr}{4\pi\epsilon_0 R^3} or \displaystyle m\frac{d^2r}{dt^2}=-\frac{qQr}{4\pi\epsilon_0 R^3} \Rightarrow \frac{d^2r}{dt^2}=-\left(\frac{qQ}{4\pi\epsilon_0 mR^3}\right)r Thus, \displaystyle \omega=2\pi f=\sqrt{\frac{qQ}{4\pi\epsilon_0 mR^3}} \Rightarrow f=\frac{1}{2\pi}\sqrt{\frac{qQ}{4\pi\epsilon_0 mR^3}} thus x=4

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Types Of Questions

7 min read

Electric Charges And Fields

- Wondering what kind of questions that can be asked in next exam? Here you go!

JEE Advanced

Single choice correct

Two identical pendulums

A

and

B

are suspended from the same point. Both are given a positive charge, with

A

having more charge than

B

. They diverge and reach equilibrium with the strings of pendulums

A

and

B

making angles

θ_{1}

and

θ_{2}

with the vertical respectively. Then:

θ_{1}

θ_{2}

θ_{1}

θ_{2}

θ_{1} = θ_{2}

the tension in A is greater than that in B

Electric Charges And Fields

Numerical Type I

Numerical Type II

Matrix Match Type I

Matrix Match Type II

Passage Type