Mix Test - 7

Given:

Charge on dipole is \(\pm 5 \mu {C}=\pm 5 \times 10^{-6} {C}\) 

Distance between the charges \(=1 {~mm}=10^{-3} {~m}\)

Dipole moment is given by:

\(P={q}(2 {a})\)

\(=5 \times 10^{-6} \times 2 \times 10^{-3}\)

\(=10^{-8} {Cm}\)

Hence, the correct option is (B).

The relative permittivity of the water is highest among the other materials given.

Permittivity describes the amount of charge needed to generate one unit of electric flux in a particular medium. Accordingly, a charge will yield more electric flux in a medium with low permittivity than in a medium with high permittivity. Thus, permittivity is the measure of a material's ability to resist an electric field.

Relative permittivity is the ratio of its absolute permittivity ‘\(\epsilon \)’ to free space (empty of matter) permittivity '\(\epsilon_0 \)’ i.e.,

\(\epsilon_{r}=\frac{\epsilon}{\epsilon_{0}}\)

Hence, the correct option is (A).

We know that:

Charge of one electron \(=1.6 \times 10^{-19}\) Coulomb.

Therefore,

Charge of \(25 \times 10^{31}\) electrons will be:

\({Q}=25 \times 10^{31} \times 1.6 \times 10^{-19}\)

\({Q}=40 \times 10^{12} {C}\)

Hence, the correct option is (C).

Let us assume that the charge \(q=\pm 10 \mu {C}=10^{-5} {C}\) is placed at a distance of \(5 {~cm}\) from the square \({ABCD}\) of each side \(10 {~cm}\). The square \({ABCD}\) can be considered as one of the six faces of a cubic Gaussian surface of each side \(10 {~cm}\).

Now, the total electric flux through the faces of the cube as per Gaussian theorem:

\(\phi=\frac{q}{\epsilon_{0}}\)

Therefore, the total electric flux through the square ABCD will be:

\(\phi_{{E}}=\frac{1}{6} \times \phi\)

\(=\frac{1}{6} \times \frac{{q}}{\epsilon_{0}}\)

\(=\frac{1}{6} \times \frac{10^{-5}}{8.854 \times 10^{-12}} \quad\) \((\because \epsilon_0 =8.854 \times 10^{-12})\)

\(=1.88 \times 10^{5} {Nm}^{2} {C}^{-1}\)

Hence, the correct option is (A).

The net electrostatic force on the charge qx by the charges q₂ and q₃ is along the positive x-direction. Therefore, the nature of force between q₁, q₂ and q₁, q₃ should be attractive. It means qx should be negative.

Obviously, due to addition of positive charge Q at (x,0), the force on −q shall increase along the positive x-axis.

Hence, the correct option is (A).

Given that:

The radius of the outer shell = b

The radius of the inner shell = a

The inner surface of the outer shell has charge +Q.

The outer surface of the inner shell has induced charge −Q.

We know that:

The potential difference between the two shells is given by,

\(V=\frac{Q}{4 \pi \epsilon_{0} a}-\frac{Q}{4 \pi \epsilon_{0} b}\)

Where \(\epsilon_{0}\) is the permittivity of free space,

\(V=\frac{Q}{4 \pi \epsilon_{0}}\left[\frac{1}{a}-\frac{1}{b}\right]\)

\(\frac{Q}{V}=\frac{4 \pi \epsilon_{0}}{\left[\frac{1}{a}-\frac{1}{b}\right]}\)

\(C=4 \pi \epsilon_{0} \left(\frac{a b}{b-a}\right)\)

Hence, the correct option is (A).

Since the particle carrying positive charge is revolving around another charge,

Electrostatic force = Centrifugal force

\(\Rightarrow\) \(\frac{1}{4 \pi \epsilon_{0}} \frac{{q}_{1} {q}_{2}}{{r}^{2}}={mr} \omega^{2}\)

\(\Rightarrow\) \(\frac{1}{4 \pi \epsilon_{0}} \frac{{q}_{1} {q}_{2}}{{r}^{2}}=\frac{4 \pi^{2} {mr}}{{T}^{2}}\)

\(\Rightarrow\) \({T}^{2}=\frac{\left(4 \pi \epsilon_{0}\right) {r}^{2}\left(4 \pi^{2} {mr}\right)}{{q}_{1} {q}_{2}}\)

\(\Rightarrow\) \({T}=4 \pi {r} \sqrt{\frac{\pi \epsilon_{0} {mr}}{{q}_{1} {q}_{2}}}\)

Hence, the correct option is (A).

Given, 

Electric field, \(E=10^{4} {NC}^{-1}\) 

Torque,

\({T}=9 \times 10^{-26} {Nm}\)

\(\theta=30^{\circ}\)

When electric dipole is placed at an angle \(\theta\) with the direction of the electric field, torque acting on the dipole is given by,

\({T}=p {E} \sin \theta\)

\(\therefore\) The electric moment of the dipole,

\( p =\frac{{T}}{{E} \sin \theta}\)

\(=\frac{9 \times 10^{-26}}{10^{4} \times \sin 30^{\circ}}\)

\(=\frac{9 \times 10^{-26}}{10^{4} \times 0.5}\)

\(=1.8 \times 10^{-19} {Cm} \)

Hence, the correct option is (A).

Given,

Thickness of the dielectric slab, \({t}=1 {~cm}=10^{-2} {~m}\)

Dielectric constant, \(\varepsilon_{{r}}={K}=5\)

Area of the plates of the capacitor, \(A=0.01 {~m}^{2}=10^{-2} {~m}^{2}\)

Distance between parallel plates of the capacitor, \({d}=2 {~cm}=2 \times 10^{-2} {~m}\)

We know that:

Capacity with air in between the plates,

\(C_{0}=\frac{\epsilon_{0} A}{d}\)

where, \(\epsilon_{0}=8.854 \times 10^{-12} \)

\(=\frac{8.85 \times 10^{-12} \times 10^{-2}}{2 \times 10^{-2}}\)

\({C}_{0}=4.425 \times 10^{-12}\) Farad

Capacity with dielectric slab in between the plates,

\({C}=\frac{\epsilon_{0} {~A}}{{~d}-{t}\left(1-\frac{1}{{~K}}\right)}\)

\(=\frac{8.85 \times 10^{-12} \times 10^{-2}}{\left(2 \times 10^{-2}\right)-10^{-2}\left(1-\frac{1}{5}\right)}\)

\(C=7.375 \times 10^{-12}\) Farad 

Increase in capacity on introduction of dielectric:

\({C}-{C}_{0}=\left(7.375 \times 10^{-12}\right)-\left(4.425 \times 10^{-12}\right)\)

\(=2.95 \times 10^{-12}\) Farad 

Hence, the correct option is (C).

Given,

Charge on an alpha particle, \(q_{1}=q_{2}=+2 {e}\)

Distance between the particles, \(r=3.2 \times 10^{-15} {~m}\)

We know that:

Charge on an electron, \(e=1.6 \times 10^{-19}\)

and, \(\frac{1}{4\pi\epsilon_0}=9 \times 10^{9}\)

Now, using coulomb's law, we get, 

Force acting on the particles is given by,

\(\begin{aligned} F &=\frac{1}{4 \pi \epsilon_{0}} \frac{q_{1} q_{2}}{r^{2}} \\ F &=\frac{9 \times 10^{9} \times 2 \times 1.6 \times 10^{-19} \times 2 \times 1.6 \times 10^{-19}}{3.2 \times 10^{-15} \times 3.2 \times 10^{-15}} {~N} \end{aligned}\)

\(F=90 {~N}\)

Hence, the correct option is (C).

Here, F = ?

Charge at \(P\),

\(q=1 \mu {C}=10^{-6} {C}\)

Charge at \(A\),

\(q_{1}=+2 \times 10^{-8} {C}\)

Charge at \(B\),

\(q_{2}=-10^{-8} {C}\)

\(A P=10 {~cm}=0.1 {~m}\)

\(B P=5 {~cm}=0.05 {~m}\)

\(\angle A P B=90^{\circ}\)

Force at \(P\) due to \(q_{1}\) charge at \(A\),

\(F_{1}=\frac{1}{4 \pi \epsilon_{0}} \frac{q_1 q}{(A P)^{2}}\), along AP produced,

\(=\frac{9 \times 10^{9} \times 2 \times 10^{-8} \times 10^{-6}}{(0.1)^{2}}\)

\(=18 \times 10^{-3} {~N}\)

Force at \(P\) due to \(q_{2}\) charge at \(B\),

\(F_{2}=\frac{1}{4 \pi \epsilon_{0}} \frac{q_2 q}{(B P)^{2}}\), along \(B P\)

\(=\frac{9 \times 10^{9} \times 10^{-8} \times 10^{-6}}{(0.05)^{2}}\)

\(=36 \times 10^{-3} {~N}\)

As angle between \(\vec{F}_{1}\) and \(\vec{F}_{2}\) is \(90^{\circ}\),

\(\therefore~\) Resultant force \(F=\sqrt{F_{1}^{2}+F_{2}^{2}}\)

\(\begin{aligned} F &=\sqrt{\left(18 \times 10^{-3}\right)^{2}+\left(36 \times 10^{-3}\right)^{2}} \\ &=18 \times 10^{-3} \times 2.236 \\ F &=4.024 \times 10^{-2} {~N} \end{aligned}\)

Hence, the correct option is (C).

Given a mass of copper slab \(=2\) g contains \(2 \times 10^{22}\) atoms.

We have to find the fraction of electrons that must be removed from sphere to give it \(+2 \mu \mathrm{C}\) charge.

Charge on nucleus of each atom \(=29 \mathrm{e}\)

\(\therefore\) Net charge on \(2 \mathrm{gm}\) sphere \(=(29 \mathrm{e}) \times\left(2 \times 10^{22}\right)=5.8 \times 10^{23} \mathrm{e} \mathrm{c}\)

\(\therefore\) No of electrons on sphere \(=5.8 \times 10^{23}\)

\(\therefore\) Number of electrons removed to give \(2 \mu \mathrm{c}\) charge \(=\frac{\mathrm{q}}{\mathrm{e}}\)

\( =\frac{2 \times 10^{-6}}{1.6 \times 10^{-19}}\)  

\(=1.25 \times 10^{13}\)

Fraction of electrons removed \(=\frac{1.25 \times 10^{13}}{\text { Total number of electrons in sphere }}\)

\(=\frac{1.25 \times 10^{13}}{29 \times 2 \times 10^{22}}\)

\(=2.16 \times 10^{-11}\)

Hence, the correct option is (D).

We know that:

A capacitor consists of two plates of a conductor and a dielectric insulator between them:

\(C=\frac{\epsilon A}{d}\)

Where

C = capacitance in farad

ϵ = Permittivity of dielectric

A = area of plate overlap in square meters

d = distance between plates in meters

If the area of the plates is halved i.e., \(\frac{{A}}{2}\) and the distance between them is halved i.e., \(\frac{d}{2}\), then new capacitance is:

\(C^{\prime}=\frac{\epsilon \frac{A}{2}}{\frac{d}{2}}\)

\(\therefore {C}^{\prime}={C}\)

Hence, the correct option is (C).

Let C be the capacitance of the infinite ladder. 

As the ladder is infinite, addition of one more element of two capacitors (1 μF and 2 μF) across the points X and Y should not change the total capacitance. 

Therefore, total capacity of the arrangement shown in figure must remain C only.

In figure, 2 μF capacitor is in series with capacitance C.

\(\therefore\) Their combined capacity \(=\frac{2 \times C}{2+C}\)

This combination is in parallel with \(1 \mu {F}\) capacitor. 

The equivalent capacity of the arrangement is:

\(1+\frac{2 C}{2+C}=C\)

or, \(C^{2}+2 C=2+3 C\)

or, \(C^{2}-C-2=0\)

\(\therefore C=2, -1\)

As capacitance cannot be negative.

\(\therefore  C=2 \mu {F}\)

Hence, the correct option is (B).

Given that, 

Charge, \((Q)=200 {nC}\)

Separation between the plates, \(({d})=5 {~mm}\)

Area of the plate, \((A)=10\) sq.cm.

Relative permeability of air \(\left(\epsilon_{r}\right)=1\)

We know that:

Capacitance, \(C=\frac{\epsilon_{0} \epsilon_{r} A}{d}\)

\(C=\frac{8.85 \times 10^{-12} \times 1 \times 10 \times 10^{-4}}{5 \times 10^{-3}}\)

\(C=1.77 {pF}\)

Also, we know that:

\(Q=C V\)

\(V=\frac{Q}{C}\)

\(=\frac{2 \times 10^{-9}}{1.77 \times 10^{-12}}\)

\(V=1130 {~V}\)

Hence, the correct option is (C).

Given,

Electric field, \({E}=9 \times 10^{4} {~N} / {C}\) 

Distance, \(r=2 \times 10^{-2} {~m}\)

Using the formula of electric field for uniformly charged wire,

\(E=\frac{\lambda}{2 \pi r \varepsilon_{0}}\)

\(\therefore \lambda={E} \cdot 2 \pi {r} \cdot \varepsilon_{0}\)

where, \(\lambda\) is a linear charge density,

and, \(\epsilon_{0}=8.854 \times 10^{-12} \)

Then,

\(\lambda=9 \times 10^{4} \times 2 \pi \times 2 \times 10^{-2} \times 8.854 \times 10^{-12}\)

\(\lambda=10 \times 10^{-6}\)

Therefore, 

Linear charge density, \( \lambda= 10 \mu C/ m \)

Hence, the correct option is (B).

Given:

Electrostatic force of repulsion, \( \mathrm{F}=3.7 \times 10^{-9} \mathrm{~N}\)

Let us say charge is \(q_{1}=q_{2}=q\)

Distance between two charges, \( r=5\)Å\(=5 \times 10^{-10} \mathrm{~m}\)

The number of electrons missing, \(n=?\)

Using Coulomb's law,

\(\mathrm{F}=\frac{1}{4 \pi \in_{0}} \frac{\mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}^{2}}\)

\(\Rightarrow 3.7 \times 10^{-9}=9 \times 10^{9} \times\frac{\mathrm{q\times  q}}{\left(5 \times 10^{-10}\right)^{2}}\)

\(\Rightarrow \mathrm{q}^{2}=\frac{3.7 \times 10^{-9} \times 25 \times 10^{-20}}{9 \times 10^{9}}\)

\(\Rightarrow \mathrm{q}^{2}=10.28 \times 10^{-38}\)

\(\Rightarrow q=3.2 \times 10^{-19}\) Coulomb

As, we know that:

\(q=\text { ne }\)

\(\therefore n=\frac{q}{e}\)

\(=\frac{3.2 \times 10^{-19}}{1.6 \times 10^{-19}}\)

\(=2\)

Therefore, the number of electrons missing is \(n=2\).

Hence, the correct option is (A).

We know that:

Permittivity of free space is given by: 

\(\epsilon_{0}=\frac{\mathrm{q}_{1} \mathrm{q}_{2}}{4 \pi \mathrm{Fr}^{2}}\)

Dimensions of \([\mathrm{F}]=\left[\mathrm{MLT}^{-2}\right]\)

Dimensions of \([\mathrm{q}]=[\mathrm{AT}]\)

Dimensions of \([\mathrm{r}]=[\mathrm{L}]\)

Thus, dimensional formula of \(\left[\epsilon_{0}\right]=\frac{[\mathrm{AT}][\mathrm{AT}]}{\left[\mathrm{MLT}^{-2}\right]\left[\mathrm{L}^{2}\right]}\)

\(\Rightarrow\left[\epsilon_{0}\right]=\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^{4} \mathrm{~A}^{2}\right]\)

Hence, the correct option is (D).

Given that:

Dipole Moment, \(P = 3 \times 10^{-3} {~cm}\) 

Electric Field Intensity, \(E = 10^{4} {NC}^{-1}\)

We know that:

In rotating the dipole from the position of stable equilibrium by an angle θ, the amount of work done is given by,

\({W}={PE}(1-\cos \theta)\)

For unstable equilibrium, \(\theta=180^{\circ}\)

\(\therefore\) \(W=P E\left(1-\cos 180^{\circ}\right) \quad\left[\because \cos 180^{\circ}=-1\right]\)

\(=2 P E\)

\(=2 \times 3 \times 10^{-3} \times 10^{4} {~J}\)

\(=60 {~J}\)

Hence, the correct option is (B).

Given that:

Separation distance, \(r=5.12 \times 10^{-15} {~m}\)

We know that:

Charge on an electron, \(e= 1.6 \times 10^{-19} \)

and, \(\frac{1}{4 \pi\epsilon_0}= 9 \times 10^{9} \)

Charge on an alpha particle is 2e.

\(\therefore\) Using Coulomb's law,

\({F}=\frac{1}{4 \pi \epsilon_{0}} \frac{{q}_{1} {q}_{2}}{{r}^{2}}\)

\(=\frac{1}{4 \pi \epsilon_{0}} \frac{2 \times 1.6 \times 10^{-19} \times 1.6 \times 10^{-19}}{\left(5.12 \times 10^{-15}\right)^{2}}\)

\(=9 \times 10^{9} \times 0.195 \times 10^{-8}\)

\(=17.5 {~N}\)

Hence, the correct option is (C).

The electric field due to any charge will be continuous, if there is no other charge in the medium. It will be discontinuous if there is a charge at the point under consideration.

Hence, the correct option is (D).

Under the static condition, the electric field inside the solid perfect conductor is zero in electrostatic equilibrium even if:

• It is isolated.
• It is charged.
• It is present in the external electrostatic field.

Now, it is known that a perfect solid conductor is an equipotential body. The potential inside the conductor is the same as the potential at the surface.

Since the surface of the solid conductor is an equipotential surface, therefore, the electric field will be perpendicular or in other words normal to the perfectly solid conductor surface.

Hence, the correct option is (C).

Given,

Diameter of the sphere \(=2.4\)

\(\therefore\) Radius of sphere, \(r=\frac{2.4}{2}=1.2 {~m}\)

Surface charge density of conducting sphere, \(\sigma=80 \times 10^{-6} {C} / {m}^{2}\)

Therefore,

Charge on sphere will be:

\(q=\sigma A=\sigma 4 \pi r^{2}\)

\(q=80 \times 10^{-6} \times 4 \times 3.14 \times(1.2)^{2}\)

\(q=1.45 \times 10^{-3} {C}\)

Then, the total electric flux leaving the surface of the sphere will be calculated using the gauss formula, i.e.,

\(\phi=\frac{q}{\varepsilon_{0}}\)

\(\phi=\frac{1.45 \times 10^{-3}}{8.854 \times 10^{-12}} \quad\) \( (\because \epsilon_{0}=8.854 \times 10^{-12}) \)

\(\phi=1.6 \times 10^{8} {Nm}^{2} / {C}\)

Hence, the correct option is (D).

We know that:

Electrostatic force \(=\frac{K Q_{1} Q_{2}}{r^{2}}\)\(\quad\)......(1)

Gravitational force \(=\frac{G m_{1} m_{2}}{r^{2}}\)\(\quad\)......(2)

On dividing the two we get,

\(\frac{\text{Electrostatic Force}}{\text{Gravitational Force}}=\frac{\frac{{KQ}_{1} \mathrm{Q}_{2}}{\mathrm{r}^{2}}}{\frac{{Gm}_{1} \mathrm{~m}_{2}}{ \mathrm{r}^{2}}}\)

\(\frac{\text{Electrostatic Force}}{\text{Gravitational Force}}=\frac{9.0 \times 10^{9} \times\left(1.6 \times 10^{-19}\right)^{2}}{6.67 \times 10^{-11} \times\left(9.1 \times 10^{-31}\right)^{2}}\)

Electrostatic Force \(=4 \times 10^{42}\) Gravitational Force 

Therefore, electrostatic force between two electrons is greater than gravitational force by a factor of \(10^{42}\).

Hence, the correct option is (B).

For infinite long line charge,

\(E=\frac{\lambda}{2 \pi \epsilon_{0} r}=\frac{2 \lambda}{4 \pi \epsilon_{0} r}\)

\(\lambda=\) linear charge density

Distance, \(r= 2 {~cm} \) 

We have,

\({E}=9 \times {10}^{4} {~N} / {C}\)

Therefore,

\(9 \times 10^{4}=\frac{2 \times \lambda \times 9 \times 10^{9}}{2 \times 10^{-2}} \quad\) \((\because \frac {1}{4 \pi \epsilon_0} =9 \times 10^{9} \text { N m}^2 / \text{C}^2 ) \)

\(\lambda=\frac{10^{4}}{10^{11}}=10^{-7} {~cm}\)

Hence, the correct option is (D).

Let, 

Effective capacitance = C 

Then:

\(\frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}\)

\(=\frac{1}{8}+\frac{1}{16}+\frac{1}{24}\)

\(=\frac{6+3+2}{48}\)

\(C=\frac{48}{11} \mu F\)

Potential difference, \(V=110 {~V}\)

We know that:

Charge, \(Q=C V\)

\(Q=\frac{48}{11} \times 10^{-6} \times 110\)

\(=480 \times 10^{-6}\)

\(=48 \times 10^{-5}\)

Since the capacitors are in series each capacitor has same charge i.e., \(48 \times 10^{-5} {C}\).

Hence, the correct option is (B).

According to gauss's law, total electric flux through a closed surface enclosing a charge is \(\frac{1}{\epsilon_{0}}\) times the magnitude of the charge enclosed.

\(\phi_{\text {electric }}=\frac{Q}{\epsilon_{0}}\)

\(\oint \vec{E} \cdot d \vec{S}=\frac{Q_{\text {in }}}{\epsilon_{0}}\)

Where:

\(\phi :\) electric flux

\(Q_{\text {in }} :\) charge enclosed the sphere

\(\epsilon_{0} :\) permittivity of space \(\left(\frac{8.85 \times 10^{-12} {C}^{2}}{{Nm}^{2}}\right)\)

\({dS} :\) surface area

Hence, the correct option is (A).

Let the charge distribution be as shown in figure:

\(\therefore V_{A}-V_{B}=\frac{q}{C_{1}}-E+\frac{q}{C_{2}}\)

or, \( \left(V_{A}-V_{B}\right)+E=q\left(\frac{1}{C_{1}}+\frac{1}{C_{2}}\right)\)

\( \left(V_{A}-V_{B}\right)+E=\frac{{q}\left({C}_{2}+C_{1} \right)}{{C}_{1} {C}_{2}}\)

\(\therefore  {q}=\frac{\left[\left({V}_{{A}}-{V}_{{B}}\right)+{E}\right] {C}_{1} {C}_{2}}{{C}_{1}+{C}_{2}}\)

Voltage across \(C_{1}\) is \(V_{1}=\frac{q}{C_{1}}\)

\(=\frac{ \left [\left(V_{A}-V_{B}\right)+E\right] C_{2}}{C_{1}+C_{2}}\) 

\(=\frac{(5+10) 2.0}{1.0+2.0}\)

\(=10\) Volt

Voltage across \(C_{2}\) is \(V_{2}=\frac{q}{C_{2}}\)

\(=\frac{ \left [\left(V_{A}-V_{B}\right)+E\right] C_{1}}{C_{1}+C_{2}}\) 

\(=\frac{(5+10) 1.0}{1.0 \times 2.0}\)

\(=5\) Volt

Hence, the correct option is (A).

Given:

Charge, \( {q}_{1}=200 \times 10^{-6} {C}=2 \times 10^{-4} {C}\)

Charge, \( {q}_{2}=500 \times 10^{-6} {C}=5 \times 10^{-4} {C}\)

Electrostatic force, \({F}=5 {gf}=5 \times 10^{-3} {~kgf}\)

\(=5 \times 10^{-3} \times 10 {~N}\)

\(=5 \times 10^{-2} {~N}\)

We have to find the distance between two charges i.e., r

Using the formula:

\({F}=\frac{1}{4 \pi \varepsilon_{0}} \frac{{q}_{1} {q}_{2}}{{r}^{2}}\),  

\(\Rightarrow 5 \times 10^{-2}=\frac{9 \times 10^{9} \times 2 \times 10^{-4} \times 5 \times 10^{-4}}{r^{2}}\)

\(\Rightarrow r=1.34 \times 10^{2} {~m}\)

Hence, the correct option is (C).

As given work is done in moving a charge of 2 coulombs across two points having a potential difference of 5 volts.

Therefore,

\({q}=2\) coulomb

\({V}=5 \) Volt

We know that:

Work Done, \( {W}={q} {V}\)

Where q is the charge flowing, W is the work done and V is the potential difference.

\({W}=2 \times 5\)

\({~W}=10\) joule

Hence, the correct option is (C).