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

By Gauss’s theorem, total electric flux over the closed surface is $$\frac{1}{{{\varepsilon _0}}}$$ times the total charges contained inside surface.
∴ Total electric flux $$ = \frac{{{\text{total charge inside cube}}}}{{{\varepsilon _0}}}$$
or $$\phi = \frac{q}{{{\varepsilon _0}}}$$

The magnitude of electric field intensity due to each part of the hemispherical surface at the centre $$'O'$$ is same.

Suppose it is $$E.$$
$$\eqalign{ & E + \frac{E}{2} + \frac{E}{2} = {E_0} \cr & {\text{or}}\,\,2E = {E_0}\,{\text{or}}\,E = \frac{{{E_0}}}{2} \cr} $$

$$\eqalign{ & \overrightarrow \tau = \overrightarrow P \times \overrightarrow E = PE\sin \theta \cr & {\text{For}}\,\,q = {90^ \circ },{\overrightarrow \tau _{\max }} = PE \cr} $$

The two charges form an electric dipole. If we take a point $$M$$ on the $$X$$-axis as shown in the figure, then the net electric field is in $$X$$-direction.
$$\therefore $$ Option (A) is incorrect. If we take a point $$N$$ on $$Y$$-axis, we find net electric field along $$+X$$ direction.The same will be true for any point on $$Y$$-axis. (B) is a correct option.
NOTE : For any point on the equatorial line of a dipole, the electric field is opposite to the direction of dipole moment.
(B) $${W_{\infty - 0}} = q\left( {{V_\infty } - {V_0}} \right) = q\left( {0 - 0} \right) = 0$$
$$\therefore $$ (C) is incorrect. The direction of dipole moment is from -ve to + ve. Therefore (D) is incorrect.

The electric flux emerging from the cube is
$$\eqalign{ & \phi = \frac{1}{{{\varepsilon _0}}} \times {\text{charge enclosed}}\,\,\left( {{q_{{\text{inside}}}}} \right) \cr & = \frac{1}{{{\varepsilon _0}}} \times {q_{{\text{inside}}}} \times {10^{ - 6}} \cr & = \frac{1}{{{\varepsilon _0}}} \times {q_{{\text{inside}}}} \times {10^{ - 6}}\,\,\,\left[ {{q_{{\text{inside}}}} = q \times {{10}^{ - 6}}} \right] \cr} $$
Since, a cube has six faces, so electric flux through each face is,
$$\phi ' = \frac{\phi }{6} = \frac{1}{{6{\varepsilon _0}}} \times q \times {10^{ - 6}} = \frac{{q \times {{10}^{ - 6}}}}{{6{\varepsilon _0}}}$$

$$\eqalign{ & E = \frac{q}{{4\pi {\varepsilon _o}{r^2}}} \Rightarrow Ar = \frac{q}{{4\pi { \in _0}{r^2}}} \cr & \Rightarrow q = 4\pi {\varepsilon _o}A{r^3} \cr} $$

$${\text{Here,}}\,\,q = 1\,C,{\varepsilon _0} = 8.85 \times {10^{ - 12}}\,{C^2}{N^{ - 1}}{m^{ - 2}}$$
Number of lines of force = Electric force
$$ = \frac{q}{{{\varepsilon _0}}} = \frac{1}{{8.85 \times {{10}^{ - 12}}}} = 1.13 \times {10^{11}}$$

$$\eqalign{ & - eE = mg \cr & \vec E = - \frac{{9.1 \times {{10}^{ - 31}} \times 10}}{{1.6 \times {{10}^{ - 19}}}} = - 5.6 \times {10^{ - 11}}\,N/C \cr} $$

$$\eqalign{ & x = \frac{1}{2}a{t^2} = \frac{1}{2}\left( {\frac{{Eq}}{m}} \right){t^2}\,{\text{and}}\,x = \frac{1}{2}\left( {\frac{{Eq}}{{2m}}} \right){{t'}^2} \cr & \therefore t' = \sqrt 2 t. \cr} $$

Choose the three coordinate axes as $$X,Y$$  and $$Z$$ and plot the charges with the given coordinates as shown. $$O$$ is the origin at which $$- 2q$$  charge is placed. The system is equivalent to two dipoles along $$x$$ and $$y$$-directions respectively. The dipole moments of two dipoles are shown in figure.

The resultant dipole moment will be directed along $$OP$$ where coordinate of $${P_{{\text{net}}}}$$ is given by $$\left( {a,a,0} \right).$$  The magnitude of resultant dipole moment is $${P_{{\text{net}}}} = \sqrt {{p^2} + {p^2}} $$
$$\eqalign{ & = \sqrt {{{\left( {qa} \right)}^2} + {{\left( {qa} \right)}^2}} \cr & = \sqrt 2 qa \cr} $$