Explanation
Hint:-Use the formula of Coulamb's Force in a medium
Explanation:-
$$\bullet$$Force of attraction between two charges in air is given by $${{F}_{a}}=\dfrac{q_1q_2}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}$$
$$\bullet$$When the air is replaced by dielectric constant K, the force of attraction between two charges becomes $${{F}_{m}}=\dfrac{q_1q_2}{4\pi K{{\varepsilon }_{0}}}=\dfrac{{{F}_{a}}}{K}$$
Thus the force is decreasing by K times
$$\textbf{Hence, the correct option is (C)}$$
Potential energy $$=\cfrac{1}{4\pi \epsilon_o}q.\cfrac{Q}{r}$$
$$|\Delta P|=\cfrac{1}{4\pi \epsilon_o}q.\cfrac{Q}{xdx}-\cfrac{1}{4\pi \epsilon_o}\cfrac{qQ}{x+dx}$$
$$lim_dx \rightarrow 0=\cfrac{1}{4\pi \epsilon_o}.\cfrac{Q.2x}{x^2-(dx)^2}$$
Where $$dx=0$$
so the answer is $$x^2$$
Three charge +4q, Q and q are placed in a straight line of length $$l$$ at points distance 0, $$\dfrac{l}{2}$$ and $$l$$ respectively. What should be the value of Q in order to make the net force on q to be zero?
A solid insulating sphere of radius a carries a net positive charge 3Q, uniformly Distributed throughout its volume. Concentric with this sphere is a conducting Spherical shell with inner radius b and outer radius c and having a net charge Q, As shown in the figure.Electric field varies with distance r from the centre as $$\left( {K = {1 \over {4\pi {\varepsilon _o}}}} \right)$$
A point charge Q is placed at origin O. Let $$\overrightarrow {{E_A}} $$,$$\overrightarrow {{E_B}} $$ and $$\overrightarrow {{E_C}} $$ represent electric fields at A, B and C respectively. If coordination of A,B and C are respectively (1,2,3) m,(1,1,-1) m and (2,2,2) m then
The chasrge density of an insulating infinity surface is $$\left( {e/\pi } \right)\;c{m^2}$$ then the field intensity at a nearby point in volt/meter will be -
1.The electric lines of force are imaginary lines the direction of filed lines for the positive charge is away from the centre and for the negative charge the field lines move towards the centre.
2. The field lines can never intersect each other because at the intersection point there can be two direction of field which is not possible.
3. The field originated from positive charge and terminate at the negative charge.
4. The lines of the force in free space is given by $$E = \frac{q}{{4\pi {\varepsilon _0}{r^2}}}$$
Given that,
Electric field $$E=500\,V/m$$
Potential $$V=3000V$$
We know that,
$$ E=\dfrac{V}{d} $$
$$ d=\dfrac{V}{E} $$
$$ d=\dfrac{3000}{500} $$
$$ d=6\,m $$
Hence, the distance is $$6\ m$$
Electric field for infinitely long wire is given by $$E=\dfrac{\lambda }{2\pi {{\varepsilon }_{0}}r}\,where\,\,\lambda =ch\arg e\,\,density$$
Since the electron is revolving then the centripetal force is balanced by the electrostatic force
$$ {{F}_{e}}={{F}_{c}} $$
$$ eE=\dfrac{m{{v}^{2}}}{r} $$
$$ \dfrac{e\lambda }{2\pi {{\varepsilon }_{0}}}=m{{v}^{2}} $$
$$ \dfrac{e\lambda }{4\pi {{\varepsilon }_{0}}}=\dfrac{1}{2}m{{v}^{2}} $$
$$ KE=\dfrac{e\lambda }{4\pi {{\varepsilon }_{0}}} $$
Given,
Electric field, $$E=20\,N{{C}^{-1}}$$
Area, $$A=10\times {{10}^{-4}}\,{{m}^{2}}$$ A
Electric Flux $$\phi =E.A\,\cos \theta =20\times 10\times {{10}^{-4}}\cos {{60}^{o}}=0.01\,Vm$$
Charge $$q={{10}^{-7}}\,C$$
The electric flux is
$$ \phi =\dfrac{q}{{{\varepsilon }_{0}}} $$
$$ \phi =\dfrac{{{10}^{-7}}}{8.85\times {{10}^{-12}}} $$
$$ \phi =0.1129\times {{10}^{5}} $$
$$ \phi =1.13\times {{10}^{4}}\,N{{m}^{2}}/C $$
Hence, the flux is $$1.13\times {{10}^{4}}\,N{{m}^{2}}/C$$
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