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Four charges equal to -Q are placed at the four corners of a square and a charge q is at its centre.  If the system is in equilibrium the value of q is
(1) $-\frac{Q}{2}(1+2\sqrt{2})$
(2) $\frac{Q}{4}(1+2\sqrt{2})$
(3) $-\frac{Q}{4}(1+2\sqrt{2})$
(4) $\frac{Q}{2}(1+2\sqrt{2})$

Two balls of same mass and carrying equal charge are hung from a fixed support of length l. At electrostatic equilibrium, assuming that angles made by each thread is small, the separation, x between the balls is proportional to:
(1) l
(2) $l^2$
(3) $l^{2/3}$
(4) $l^{1/3}$

A solid conducting sphere of radius a has a net positive charge 2Q.  A conducting spherical shell of inner radius b and outer radius c is concentric with the solid sphere and has a net charge -Q.  The surface charge density on the inner and outer surfaces of the spherical shell will be 

(1) $-\frac{2Q}{4{\mathrm{\pi b}}^2},\frac{Q}{4{\mathrm{\pi c}}^2}$
(2) $-\frac{Q}{4{\mathrm{\pi b}}^2},\frac{Q}{4{\mathrm{\pi c}}^2}$
(3) $0,\frac{Q}{4{\mathrm{\pi c}}^2}$
(4) None of the above

Two pith balls carrying equal charges are suspended from a common point by strings of equal length.  The equilibrium separation between them is r.  Now the strings are rigidly clamped at half the height.  The equilibrium separation between the balls now become

(1) $\left(\frac{r}{\sqrt[3]{2}}\right)$
(2) $\left(\frac{2r}{\sqrt{3}}\right)$
(3) $\left(\frac{2r}{3}\right)$
(4) ${\left(\frac{r}{\sqrt{2}}\right)}^2$

Two particles A and B having equal charges are placed at a distance d apart.  A third charged particle placed on the perpendicular bisection of AB at distance x.  The third particle experiences maximum force when 
(1) $x=\frac{d}{\sqrt{2}}$
(2) $x=\frac{d}{2}$
(3) $x=\frac{d}{2\sqrt{2}}$
(4) $x=\frac{d}{3\sqrt{2}}$

 A spherically symmetric charge distribution is characterised by a charge density having the following variations:
$\rho \left(r\right)={\rho }_0\left(1-\frac{r}{R}\right) for r<R$
$\rho \left(r\right)=0 for r\ge R$
Where r is the distance from the centre of the charge distribution and ${\rho }_0$is a constant.  The electric field at an internal point is -
(1) $\frac{{\rho }_0}{4{\epsilon }_0}\left(\frac{r}{3}-\frac{r^2}{4R}\right)$
(2) $\frac{{\rho }_0}{{\epsilon }_0}\left(\frac{r}{3}-\frac{r^2}{4R}\right)$
(3) $\frac{{\rho }_0}{3{\epsilon }_0}\left(\frac{r}{3}-\frac{r^2}{4R}\right)$
(4)$\frac{{\rho }_0}{12{\epsilon }_0}\left(\frac{r}{3}-\frac{r^2}{4R}\right)$

Two point dipoles of dipole moment $\overrightarrow{P_1} and \overrightarrow{P_2}$ are at a distance x from each other and $\overrightarrow{P_1} \parallel \overrightarrow{P_2}$.  The force between the dipoles is:
(1) $\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{4p_1{p}_2}{x^4}$
(2) $\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{3p_1{p}_2}{x^4}$
(3) $\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{6p_1{p}_2}{x^4}$
(4) $\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{8p_1{p}_2}{x^4}$

Point charge q moves from point P to point S along the path PQRS (as shown in fig.) in a uniform electric field E pointing co-parallel to the positive direction of X-axis.  The coordinates of the points P, Q, R and S are (a, b, 0), (2a, 0, 0), (a, -b, 0) and (0, 0, 0) respectively.  The workdone by the field in the above case is given by the expression

(1) qEa
(2) -qEa
(3) qEa$\sqrt{2}$
(4) qE$\sqrt{\left[{\left(2a\right)}^2+b^2\right]}$

In the figure the net electric flux through the area A is $\varphi =\overrightarrow{E}·\overrightarrow{A}$ when the system is in air.  On immersing the system in water the net electric flux through the area
 
(1) becomes zero
(2) remains same
(3) increases
(4) decreases

A thin semi-circular ring of radius r has a positive charge q distributed uniformly over it.  The net field $\overrightarrow{E}$ at the centre O is 

(1) $\frac{q}{4{\mathrm{\pi }}^2{\mathrm{\epsilon }}_0{\mathrm{r}}^2}\hat{j}$
(2) $-\frac{q}{4{\mathrm{\pi }}^2{\mathrm{\epsilon }}_0{\mathrm{r}}^2}\hat{j}$
(3) $-\frac{q}{2{\mathrm{\pi }}^2{\mathrm{\epsilon }}_0{\mathrm{r}}^2}\hat{j}$
(4) $\frac{q}{2{\mathrm{\pi }}^2{\mathrm{\epsilon }}_0{\mathrm{r}}^2}\hat{j}$

A positive charge +Q is fixed at a point A.  Another positively charged particle of mass m and charge +q is projected from a point B with velocity u as shown in the figure.  The point B is at the large distance from A and at distance d from the line AC.  The initial velocity is parallel to the line AC.  The point C is at very large distance from A.  The minimum distance (in meter) of +q from +Q during the motion is $d(1+\sqrt{A})$.  Find the value of A. $\left[Take Qq=4{\mathrm{\pi \epsilon }}_0{\mathrm{mu}}^2\mathrm{d} \mathrm{and} \mathrm{d}=\left(\sqrt{2}-1\right) \mathrm{meter}\right]$

(1) 3
(2) 2
(3) 4
(4) 5

The electric field intensity at the centre of a uniformly charged hemispherical shell is $E_0$.  Now two portions of the hemisphere are cut from either side, and the remaining portion is shown in fig.  If $\alpha =\beta =\pi /3$, then the electric field intensity at the centre due to the remaining portion is 

(1) $E_0/3$
(2) $E_0/6$
(3) $E_0/2$
(4) information insufficient

An electrostatic field line leaves at angle $\alpha$ from point charge $q_1$, and terminates at point charge $-q_2$ at an angle $\beta$ (shown in figure).  Then the relationship between $\alpha$ and $\beta$ is 

(1) $q_1 {\sin }^2\alpha =q_2 {\sin }^2\beta$
(2) $q_1 {\sin }^2\frac{\alpha }{2}=q_2 {\sin }^2\frac{\beta }{2}$
(3) $q_1 \tan \alpha =q_2 \tan \beta$
(4) $q_1 \cos \alpha =q_2 \cos \beta$

A uniform charged and infinitely long line having a linear charge density $\lambda$ is placed at a normal distance y from a point O.  Consider an imaginary sphere of radius R with O as centre and $R>y$.  Electric flux through the surface of the sphere is 

(1) zero
(2) $\frac{2\lambda R}{{\epsilon }_0}$
(3) $\frac{2\lambda \sqrt{R^2-y^2}}{{\epsilon }_0}$
(4) $\frac{\lambda \sqrt{R^2+y^2}}{{\epsilon }_0}$ 

Flux passing through the shaded surface ( ABDE) of a sphere  when a point charge q is placed at the centre is (radius of the sphere is R) -
 
(1) $q/{\epsilon }_0$
(2) $q/2{\epsilon }_0$
(3) $q/4{\epsilon }_0$
(4) zero