1.
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
A.
$$\frac{q}{{4{\pi ^2}{\varepsilon _0}{r^2}}}\hat j$$
B.
$$ - \frac{q}{{4{\pi ^2}{\varepsilon _0}{r^2}}}\hat j$$
C.
$$ - \frac{q}{{2{\pi ^2}{\varepsilon _0}{r^2}}}\hat j$$
Let us consider a differential element $$dl.$$ charge on this element.
$$\eqalign{
& dq = \left( {\frac{q}{{\pi r}}} \right)dl \cr
& = \frac{q}{{\pi r}}\left( {rd\theta } \right)\left( {\because dl = rd\theta } \right) \cr
& = \left( {\frac{q}{\pi }} \right)d\theta \cr} $$
Electric field at $$O$$ due to $$dq$$ is
$$dE = \frac{1}{{4\pi { \in _0}}} \cdot \frac{{dq}}{{{r^2}}} = \frac{1}{{4\pi { \in _0}}} \cdot \frac{q}{{\pi {r^2}}}d\theta $$
The component $$dE\cos \theta $$ will be counter balanced by another element on left portion. Hence resultant field at $$O$$ is the resultant of the component $$dE\sin \theta $$ only.
$$\eqalign{
& \therefore E = \int d E\sin \theta = \int\limits_0^\pi {\frac{q}{{4{\pi ^2}{r^2}{ \in _0}}}} \sin \theta d\theta \cr
& = \frac{q}{{4{\pi ^2}{r^2}{ \in _0}}}\left[ { - \cos \theta } \right]_0^\pi = \frac{q}{{4{\pi ^2}{r^2}{ \in _0}}}\left( { + 1 + 1} \right) \cr
& = \frac{q}{{2{\pi ^2}{r^2}{ \in _0}}} \cr} $$
The direction of $$E$$ is towards negative $$y$$-axis.
$$\therefore \overrightarrow E = - \frac{q}{{2{\pi ^2}{r^2}{ \in _0}}}\hat j$$
2.
Charges are placed on the vertices of a square as shown. Let $$\overrightarrow E $$ be the electric field and $$V$$ the potential at the centre. If the charges on $$A$$ and $$B$$ are interchanged with those on $$D$$ and $$C$$ respectively, then
A.
$$\overrightarrow E $$ changes, $$V$$ remains unchanged
B.
$$\overrightarrow E $$ remains unchanged, $$V$$ changes
C.
both $$\overrightarrow E $$ and $$V$$ change
D.
$$\overrightarrow E $$ and $$V$$ remain unchanged
Answer :
$$\overrightarrow E $$ changes, $$V$$ remains unchanged
As shown in the figure, the resultant electric fields before and after interchanging the charges will have the same magnitude, but opposite directions.
Also, the potential will be same in both cases as it is a scalar quantity.
3.
Two point dipoles $$p\hat k$$ and $$\frac{p}{2}\hat k$$ are located at $$\left( {0,0,0} \right)$$ and $$\left( {1m,0,2m} \right)$$ respectively. The resultant electric field due to the two dipoles at the point $$\left( {1m,0,0} \right)$$ is
The given point is at axis of $$\frac{{\vec p}}{2}$$ dipole and at equatorial line of $${\vec p}$$ dipole so that field at given point is $${{\vec E}_1} + {{\vec E}_2}$$
$$\eqalign{
& {{\vec E}_1} = \frac{{2k\left( {\frac{p}{2}} \right)}}{{{2^3}}} = \frac{{Kp}}{8}\left( { + \hat k} \right) \cr
& {{\vec E}_2} = \frac{{Kp}}{1}\left( { - \hat k} \right) \cr
& {{\vec E}_1} + {{\vec E}_2} = - \frac{7}{8}Kp\left( { - \hat k} \right) = - \frac{{7p}}{{32\pi { \in _0}}}\hat k \cr} $$
4.
A thin spherical shell of radius $$R$$ has charge $$Q$$ spread uniformly over its surface. Which of the following graphs most closely represents the electric field $$E\left( r \right)$$ produced by the shell in the range $$0 \leqslant r < \infty ,$$ where $$r$$ is the distance from the centre of the shell?
The electric field inside a thin spherical shell of radius $$R$$ has charge $$Q$$ spread uniformly over its surface is zero.
Outside the shell the electric field is $$E = k\frac{Q}{{{r^2}}}.$$
These characteristics are represented by graph (A).
5.
In a uniformly charged sphere of total charge $$Q$$ and radius $$R,$$ the electric field $$E$$ is plotted as function of distance from the centre, The graph which would correspond to the above will be:
6.
A charged ball $$B$$ hangs from a silk thread $$S,$$ which makes an angle $$\theta $$ with a large charged conducting sheet $$P,$$ as shown in the figure. The surface charge density $$\sigma $$ of the sheet is proportional to
7.
Two very long line charges of uniform charge density $$ + \lambda $$ and $$ - \lambda $$ are placed along same line with the separation between the nearest ends being $$2a,$$ as shown in figure. The electric field intensity at point $$O$$ is
The field at $$O$$ due to small element $$dx$$ is
$$dE = \frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{\lambda dx}}{{{x^2}}}$$
Hence, due to one wire,
$$\eqalign{
& {E_1} = \int\limits_a^\infty {\frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{\lambda dx}}{{{x^2}}}} \cr
& {E_1} = \frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{\lambda }{a}\,{\text{towards}}\,{\text{left}}{\text{.}} \cr} $$
Electric field at $$O$$ due to other wire,
$${E_2} = \frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{\lambda }{a}\,{\text{towards}}\,{\text{left}}$$
$$\therefore $$ Net field at $$O$$ is $$E = 2 \times \frac{1}{{4\pi {\varepsilon _0}}}\frac{\lambda }{a} = \frac{\lambda }{{2\pi {\varepsilon _0}a}}$$
8.
A hollow insulated conducting sphere is given a positive charge of $$10\,\mu C.$$ What will be the electric field at the centre of the sphere if its radius is $$2\,m$$ ?
Charge resides on the outer surface of a conducting hollow or solid sphere of radius $$R$$ (say). We consider a spherical surface of radius $$r < R.$$
By Gauss’ theorem, $$\sum {E \cdot ds = \frac{{{q_{{\text{inside}}}}}}{{{\varepsilon _0}}}} $$
or $$E \times 4\pi {r^2} = \frac{1}{{{\varepsilon _0}}} \times {q_{{\text{inside}}}}$$
and we know that $${q_{{\text{inside}}}} = 0$$
So, $$E = 0$$
i.e. electric field inside a hollow sphere is zero.
9.
Consider an electric field $$\vec E = {E_0}\hat x$$ where $${E_0}$$ is a constant. The flux through the shaded area (as shown in the figure) due to this field is
According to Gauss's law, the electric flux through a closed surface is equal to $$\frac{1}{{{\varepsilon _0}}}$$ times the net charge enclosed by the surface.
As the charge enclosed $$ = \frac{q}{8}$$
So, electric flux $$ = \frac{{{q_{{\text{enclosed}}}}}}{{{\varepsilon _0}}} \Rightarrow \phi = \frac{q}{{8{\varepsilon _0}}}$$