Three identical dipoles are arranged as shown below. What will be the net electric field at $P\left( {k = \frac{1}{{4\pi {\varepsilon _0}}}} \right)$

Point $P$ lies at equatorial positions of dipole 1 and 2 and axial position of dipole 3 Hence field at $P$ due to dipole 1 ${E_1} = \frac{{k.p}}{{{x^3}}}$ [towards left] due to dipole 2 ${E_2} = \frac{{k.p}}{{{x^2}}}$ [towards left] Due to dipole 3 ${E_3} = \frac{{k.\left( {2p} \right)}}{{{x^3}}}$[towards right] So net field at $P$ will be zero

A sphere of $4\;cm$ radius is suspended within a hollow sphere of $6\;cm$ radius. The inner sphere is charged to potential 3 $e.s.u.$ and the outer sphere is earthed. The charge on the inner sphere is

Suppose charge on inner sphere is $ + Q$ as shown. Potential on inner sphere $V = \frac{Q}{4} - \frac{Q}{6}$ $ \Rightarrow 3 = Q\left( {\frac{1}{4} - \frac{1}{6}} \right) \Rightarrow Q = 36\;e.s.u$

The correct Option is "$36\;e.s.u.$"

A cylindrical capacitor has charge $Q$ and length $L.$ If both the charge and length of the capacitor are doubled, by keeping other parameters fixed, the energy stored in the capacitor

Capacitance of a cylindrical capacitor $ = \frac{{2\pi {\varepsilon _0}L}}{{{\rm{ln\;}}\left( {b/a} \right)}}$ Energy stored in the capacitor $\frac{1}{2}\frac{{{Q^2}}}{C} = \frac{1}{2}\frac{{{Q^2}\;{\rm{ln\;}}\left( {b - a} \right)}}{{2\pi {\varepsilon _0}L}} = \left[ {{\rm{const}}} \right]\frac{{{Q^2}}}{L}$ If the charge is doubled and length is doubled, $\left[ {const} \right]\frac{{{Q^{'2}}}}{{L'}} = \frac{4}{2}\left( {\frac{{{Q^2}}}{L}} \right) = 2$ times the energy

The correct Option is "Increases two times"

The electric potential $V$ at any point $\left( {x,y,z} \right)$, all in meters in space is given by $V = 4{x^2}\;volt.$ The electric field at the point $\left( {1,0,2} \right)$ in $volt/meter$, is

$\vec E = - \frac{{dV}}{{dx}}\hat i = - 8x\hat i\;{\rm{volt}}/{\rm{meter}}$ ${E_{\left( {1,0,2} \right)}} = - 8\hat i\;V/m$

The correct Option is "8 along negative $X$-axis"

$N$ identical spherical drops charged to the same potential $V$ are combined to form a big drop. The potential of the new drop will be

If the drops are conducting, then $\frac{4}{3}\pi {R^3} = N\left( {\frac{4}{3}\pi {r^3}} \right) \Rightarrow R = {N^{1/3}}r.$. Final charge $Q = Nq$ So final potential $V' = \frac{Q}{R} = \frac{{Nq}}{{{N^{1/3}}r}} = V \times {N^{2/3}}$

The correct Option is "$V \times {N^{2/3}}$"

The capacitance of an air capacitor is $15\mu F$ the separation between the parallel plates is $6mm.$ A copper plate of $3mm$ thickness is introduced symmetrically between the plates. The capacitance now becomes

By using ${C_{air}} = \frac{{{\varepsilon _0}A}}{d},\;{C_{medium}} = \frac{{{\varepsilon _0}A}}{{d - t + \frac{t}{K}}}$ For $K = \infty \;\;{C_{medium}} = \frac{{{\varepsilon _0}A}}{{d - t}}$ $ \Rightarrow \frac{{{C_m}}}{{{C_a}}} = \frac{d}{{d - t}} \Rightarrow \frac{{{C_m}}}{{15}} = \frac{6}{{6 - 3}} \Rightarrow {C_m} = 30\mu \;F$

The correct Option is "$30\mu F$"

What is the equivalent capacitance between $A$ and $B$ in the given figure (all are in farad)

The given circuit can be simplified as follows Hence equivalent capacitance between $A$ and $B$ $\frac{1}{{{C_{AB}}}} = \frac{1}{{12}} + \frac{1}{{20/3}} + \frac{1}{{16}} \Rightarrow {C_{AB}} = \frac{{240}}{{71}}F$

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