Q: A very long, solid insulating cylinder with radius R has a cylindrical hole with radius a bored along its entire length. The axis of the hole is a distance b from the axis of the cylinder, where a < b < R (as shown in figure). The solid material of the cylinder has a uniform has a uniform volume charge density \rho . Find the magnitude and direction of the electric field inside the hole, and show that this is uniform over the entire hole.

\vec {E}_{whole} = \vec {E}_{cylinder} - \vec {E}_{hole} \vec {E}_{whole} = \dfrac {\rho \vec {r}}{2\epsilon_{0}} - \dfrac {\rho \vec {S}}{2\epsilon_{0}} = \dfrac {\rho}{2\epsilon_{0}} (\vec {r} - \vec {s}) = \dfrac {\rho}{2\epsilon_{0}} \vec {b} Note that \vec {E}_{whole} is uniform within the hole.

Q: A dielectric in the form of a sphere is introduced into a homogeneous electric field. A, B and C are three points as shown in Fig. Then

The dielectric gets polarized as shown. Induced charges will also contribute in electric field.So, intensity at point A and C will increase and at B intensity will decrease.

Question 1

An electron is moving with constant velocity along    x  -axis. If a uniform electric field is applied along   y  -axis, then its path in the  x-y   plane will be :

Accepted Answer

If a uniform electric field is applied in  y -direction , then a force acts in the  -y  direction for the moving electron. Thus the case is thus of a projectile and the path is a parabola.

Question 2

A charge  Q  is kept at the corner of a cube. Electric flux passing through one of those faces not touches that charge is:-

Accepted Answer

As we know that, total flux due to a charge enclosed in a region is given as  \phi=\dfrac{Q}{\varepsilon_0}  (Gauss's law)Here, the charge is on a corner of each of the 8 cubes.Hence, an equal amount of flux will spread outwards through each of the cubes due to symmetry.Hence,  the magnetic flux through each cube is given as  \phi=\dfrac{Q}{8\varepsilon_0} .Since three faces don't have any effect, the flux will be divided into three parts again.Therefore, total flux will  become,   \phi=\dfrac{Q}{24\varepsilon_0} The correct option A.

Question 3

Three concentric metallic spherical shells of radii

R, 2 R, 3 R

, are given charges

Q_{1}, Q_{2}, Q_{3}

, respectively. It is found that the surface charge densities on the outer surfaces of the shells are equal. Then, the ratio of the charges given to the shells,

Q_{1} : Q_{2} : Q_{3}

, is :

A

1 : 2 : 3

B

1 : 3 : 5

C

1 : 4 : 9

D

1 : 8 : 18

View solution

Accepted Answer

In metallic shell , all charge will appear on the outer surface of  outer shell and charge inside each is zero.The charge distributions among the shells will occur in such a way that the charge inside each shell is zero and total charge will appear on the outer surface of outer shell. (shown in fig below)Let surface density of charges of outer surfaces of each shell is  \sigma .now,  \displaystyle Q_1=\sigma.4\pi R^2 , \displaystyle Q_1+Q_2=\sigma.4\pi (2R)^2 \Rightarrow Q_2=\sigma.4\pi 4R^2-Q_1=\sigma.4\pi 4R^2-\sigma.4\pi R^2 =3(\sigma.4\pi R^2)  and \displaystyle Q_1+Q_2+Q_3=\sigma.4\pi (3R)^2   \Rightarrow Q_3= \sigma.4\pi 9R^2-(Q_1+Q_2)=\sigma.4\pi 9R^2-\sigma.4\pi 4R^2=5(\sigma.4\pi R^2) thus,  Q_1:Q_2:Q_3=1:3:5

Question 4

Two point charges +q and +9q are placed at (-a, 0) and (+a, 0). Take electric field intensity to be positive if it is along positive x-direction. The variation of the electric field intensity as one moves along the x-axis is :

Accepted Answer

E=\dfrac{F}{q}   In the middle of two charges  E \propto(\dfrac{1}{x^2} - \dfrac{9}{(r-x)^2})  where r is the separation between the two point charges and x is the distance from point charge of value +q ;Now, at  x= 0  \ or  \ x =r  , value of E is infinity ;while at  x = \dfrac{r}{2}  , it comes out to be negative  ;In the left or right of the charges i.e not in the middle along the x axis E\propto (\dfrac{1}{x^2} + \dfrac{9}{(r+x)^2}) or   (\dfrac{1}{(r+x)^2} + \dfrac{9}{(x)^2})  So, on analyzing, we can see the correct plot for the given expression will be the plot as mentioned in the option  (A)

Question 5

A very long, solid insulating cylinder with radius  R  has a cylindrical hole with radius  a  bored along its entire length. The axis of the hole is a distance  b  from the axis of the cylinder, where  a < b < R  (as shown in figure). The solid material of the cylinder has a uniform has a uniform volume charge density  \rho . Find the magnitude and direction of the electric field inside the hole, and show that this is uniform over the entire hole.

Accepted Answer

\vec {E}_{whole} = \vec {E}_{cylinder} - \vec {E}_{hole} \vec {E}_{whole} = \dfrac {\rho \vec {r}}{2\epsilon_{0}} - \dfrac {\rho \vec {S}}{2\epsilon_{0}} = \dfrac {\rho}{2\epsilon_{0}} (\vec {r} - \vec {s}) = \dfrac {\rho}{2\epsilon_{0}} \vec {b} Note that  \vec {E}_{whole}  is uniform within the hole.

Question 6

A dielectric in the form of a sphere is introduced into a homogeneous electric field.  A, B  and  C  are three points as shown in Fig. Then

Accepted Answer

The dielectric gets polarized as shown. Induced charges will also contribute in electric field.So, intensity at point  A  and  C  will increase and at  B  intensity will decrease.

Electric Charges And Fields