Electric Field MCQ Questions & Answers in Electrostatics and Magnetism | Physics

$$\eqalign{ & {E_1} = \frac{1}{{4\pi { \in _0}}} \cdot \frac{Q}{{{R^2}}}; \cr & {E_2} = \frac{1}{{4\pi { \in _0}}} \cdot \frac{{2Q}}{{{R^2}}};{E_3} = \frac{1}{{4\pi { \in _0}}} \cdot \frac{{\frac{Q}{2}}}{{{R^2}}} \cr} $$
Clearly $${E_2} > {E_1} > {E_3}$$
where $${\frac{Q}{2}}$$ is the charge enclosed in a sphere of radius $$R$$ concentric with the given sphere.
$$\left[ {\frac{{4Q}}{{\frac{4}{3}\pi {{\left( {2R} \right)}^3}}} = \frac{{Q'}}{{\frac{4}{3}\pi {R^3}}}} \right]$$

Applying Gauss's law
$$\eqalign{ & \oint_S {\overrightarrow E .\overrightarrow {ds} = \frac{Q}{{{ \in _0}}}} \cr & \therefore E \times 4\pi {r^2} = \frac{{Q + 4\pi a{r^2} - 4\pi A{a^2}}}{{{ \in _0}}} \cr & \rho = \frac{{dr}}{{dv}} \cr & Q = {\rho ^4}\pi {r^2} \cr & Q = \int\limits_a {\frac{A}{r}4\pi {r^2}dr = 4\pi A\left[ {{r^2} - {a^2}} \right]} \cr & E = \frac{1}{{4\pi { \in _0}}}\left[ {\frac{{Q - 4\pi A{a^2}}}{{{r_0}}} + 4\pi A} \right] \cr} $$
For $$E$$ to be independent of $$'r'$$
$$\eqalign{ & Q - 2\pi A{a^2} = 0 \cr & \therefore A = \frac{Q}{{2\pi {a^2}}} \cr} $$

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.
Since, $$q$$ is the charge enclosed by the surface, then electric flux, $$\phi = \frac{q}{{{\varepsilon _0}}}$$
If charge $$q$$ is placed at a corner of cube, it will be divided into 8 such cubes. Therefore, electric flux through the cube is $$\phi ' = \frac{1}{8}\left( {\frac{q}{{{\varepsilon _0}}}} \right)$$

The two forces acting on the proton just after the release are shown in the figure. In this situation
$$\eqalign{ & qE = mg\,\,\,\,\,\,\left[ {\therefore \theta = {{45}^ \circ }} \right] \cr & \therefore q\left( {\frac{V}{d}} \right) = mg \cr} $$

$$\therefore V = \frac{{mgd}}{q} = \frac{{1.67 \times {{10}^{ - 27}} \times 10 \times {{10}^{ - 2}}}}{{1.6 \times {{10}^{ - 19}}}} = {10^{ - 9}}V$$

Since $$\tau = pE\sin \theta $$   on decreasing the distance between the two charges, and on decreasing angle $$\theta $$ between the dipole and electric field, $$\sin \theta $$  decreases therefore torque decreases.

Field lines originate perpendicular from positive charge and terminate perpendicular at negative charge. Further this system can be treated as an electric dipole.

The electric field due to $$A$$ and $$B$$ at $$O$$ are equal and opposite producing a resultant which is zero. The electric field at $$O$$ due to $$C$$ is
$$E = \frac{1}{{4\pi { \in _0}}}\frac{{\frac{{2q}}{3}}}{{{R^2}}} = \frac{q}{{6\pi { \in _0}{R^2}}}.$$
$$\therefore $$ Option [A] is not correct. The electric potential at $$O$$ is

$${V_O} = K\left[ {\frac{{\frac{{ + q}}{3}}}{R}} \right] + K\left[ {\frac{{\frac{{ + q}}{3}}}{R}} \right] + K\left[ {\frac{{\frac{{ - 2q}}{3}}}{R}} \right] = 0$$
$$\therefore $$ Option [D] is wrong
In $$\Delta ABC\frac{{AC}}{{AB}} = \sin {30^ \circ } \Rightarrow AC = \frac{{AB}}{2} = R$$
Also $$\frac{{BC}}{{AB}} = \sin {60^ \circ } \Rightarrow BC = \frac{{\sqrt 3 AB}}{2} = \sqrt 3 R$$
Potential energy of the system
$$\eqalign{ & K\left[ {\frac{{\left( {\frac{q}{3}} \right)\left( {\frac{2}{3}} \right)}}{{2R}}} \right] + K\left[ {\frac{{\left( {\frac{q}{3}} \right)\left( {\frac{{ - 2q}}{3}} \right)}}{R}} \right] + K\left[ {\frac{{\left( {\frac{q}{3}} \right)\left( {\frac{{ - 2q}}{3}} \right)}}{{\sqrt 3 R}}} \right] \cr & = \frac{{k{q^2}}}{{9R}}\left[ {\frac{1}{2} - 2 - \frac{2}{{\sqrt 3 }}} \right] \ne 0 \cr} $$
$$\therefore $$ Option [B] is wrong
Magnitude of force between $$B$$ and $$C$$ is
$$F = \frac{1}{{4\pi { \in _0}}}\frac{{\left( {\frac{{2q}}{3}} \right)\left( {\frac{q}{3}} \right)}}{{{{(\sqrt 3 R)}^2}}} = \frac{{{q^2}}}{{54\pi { \in _0}{R^2}}}$$

Take $$PO$$ as the $$x$$-axis and $$PA$$ as the $$y$$-axis. Consider two elements $$EF$$ and $$E'F'$$  of width $$d\theta $$ at angular distance $$\theta $$ above and below $$PO,$$  respectively. The magnitude of the fields at $$P$$ due to either element is
$$dE = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{\frac{{rd\theta \times Q}}{{\left( {\frac{{\pi r}}{2}} \right)}}}}{{{r^2}}} = \frac{Q}{{2{\pi ^2}{\varepsilon _0}{r^2}}}d\theta $$
Resolving the fields, we find that the components along $$PO$$  sum up to zero, and hence the resultant field is along $$PB.$$  Therefore, field at $$P$$ due to pair of elements is $$2d\,E\sin \theta $$
$$\eqalign{ & E = \int_0^{\frac{\pi }{2}} {2d\,E\sin \theta } \cr & = 2\int_0^{\frac{\pi }{2}} {\frac{Q}{{2\pi {\varepsilon _0}{r^2}}}\sin \theta d\theta } = \frac{Q}{{{\pi ^2}{\varepsilon _0}{r^2}}} \cr} $$

$$w = \vec F.\vec s = Eq\hat i.\left( { - a\hat i - b\hat j} \right) = - Eqa$$

According to Gauss’s law total electric flux through a closed surface is $$\frac{1}{{{\varepsilon _0}}}$$ times the total charge inside that surface.
Electric flux, $${\phi _E} = \frac{q}{{{\varepsilon _0}}}$$
Charge on $$\alpha $$-particle$$ = 2e$$
$$\therefore {\phi _E} = \frac{{2e}}{{{\varepsilon _0}}}$$