Electric Charges and Fields Test

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Electric Charges and Fields Test - 21

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Question 1
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A cone of base radius $$R$$ and height $$h$$ is located in a uniform electric field $$\overrightarrow{E}$$ parallel to its base. The electric flux entering the cone is:

A
$$4 E h R$$

B
$$\displaystyle \dfrac{1}{2}$$ $$E h R$$

C
$$2 E h R$$

D
$$E h R$$

Solution

We know that,
$$\displaystyle \phi = E.A Cos \theta$$
Flux crossing through the area of cone parallel to the electric field which is nothing but a right angle triangle having base $$2R$$ and height $$h$$
therefore, $$\displaystyle \phi= {EhR}$$
Question 2
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Two spherical conductors B and C having equal radii and carrying equal charges in them repel each other with a force F when kept apart at some distance. A third spherical conductor having same radius as that of B but uncharged, is brought in contact with B, then brought in contact with C and finally removed away from both. The new force of repulsion between B and C is:

A
$$\displaystyle \dfrac{\mathrm{F}}{4}$$

B
$$\displaystyle \dfrac{3\mathrm{F}}{4}$$

C
$$\displaystyle \dfrac{\mathrm{F}}{8}$$

D
$$\displaystyle \dfrac{3\mathrm{F}}{8}$$

Solution

Let initially B and C have charges $$Q$$ each and distance between them be $$r$$.
$$F= \dfrac{Q^2}{4\pi \epsilon_o r^2}$$
Now when another uncharged spherical conductor A of same radius is brought in contact with B, then they will share equal charge i.e., A and B will have charge
$$q=\dfrac{Q_1+Q_2}{2}$$ each.

i.e $$q=\dfrac{Q}{2}$$

Similarly, now when A is brought in contact with C, then both will have equal charge let say $$q'.$$
$$q' = \dfrac{Q + \dfrac{Q}{2}}{2} = \dfrac{3Q}{4}$$
Force between B and C, $$F' = \dfrac{\dfrac{Q}{2} \times \dfrac{3Q}{4}}{4\pi \epsilon_o r^2} = \dfrac{3F}{8}$$
Question 3
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The magnitude of the average electric field normally present in the atmosphere just above the surface of the Earth is about $$150 N/C$$, directed inward towards the center of the Earth. This gives the total net surface charge carried by the Earth to be : [Given $$\epsilon_0=8.85\times 10^{-12} C^2/N-m^2, R_E=6.4\times 10^6m]$$

A
$$+660 kC$$

B
$$-660 kC$$

C
$$-680 kC$$

D
$$+680 kC$$

Solution

From Gauss' Law,
the electric field entering the Earth's atmosphere is E. The flux of this field is,
$$ \int E.ds $$ = $$ E \times 4 \pi {R}^{2} $$
This should be equal to the charge enclosed inside the Earth divided by $$ \epsilon $$
Hence,
$$ E \times 4 \pi {R}^{2} $$ = $$ \dfrac{q}{\epsilon} $$
Hence, putting back the value of $$3$$ knowns, the value of $$q = -680 \times {10}^{3} C $$
Question 4
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Consider two charged metallic spheres $${S}_{1}$$ and $${S}_{2}$$ of radii $${R}_{1}$$ and $${R}_{2}$$ respectively. The electric fields $${E}_{1}$$ (on $${S}_{1}$$) and $${E}_{2}$$ (on $${S}_{2}$$) on their surfaces are such that $${ E }_{ 1 }/{ E }_{ 2 }={ R }_{ 1 }/{ R }_{ 2 }$$. Then the ratio $${V}_{1}$$ (on $${S}_{1}$$)/$${V}_{2}$$ (on $${S}_{2}$$) of the electrostatic potentials on each sphere is:

A
$${ \left( { R }_{ 1 }/{ R }_{ 2 } \right) }^{ 2 }$$

B
$${ R }_{ 1 }/{ R }_{ 2 }$$

C
$${ \left( \cfrac { { R }_{ 1 } }{ { R }_{ 2 } } \right) }^{ 3 }$$

D
$$\left( { R }_{ 2 }/{ R }_{ 1 } \right) $$

Solution

Let charge $$S_1 = Q_1$$

Charge on $$S_2 = Q_2$$

For $$S_1, E_1 = \dfrac{1}{4\pi \epsilon_0} \dfrac{Q_1}{R_1^2}$$

$$\Rightarrow Q_1 = 4\pi \epsilon_0 E_1 R_1^2$$ ....(i)

$$V_1 = \dfrac{1}{4\pi \epsilon_0} \dfrac{Q_1}{R_1}$$

$$\Rightarrow Q_1 = 4\pi \epsilon_0 V_1R_1$$ ...(ii)

Similarly for $$S_2 $$ $$Q_2 = 4\pi E_0 E_2R_2^2$$ ....(iii)

$$Q_2 = 4\pi \epsilon_0V_2R_2$$ ......(iv)

$$\dfrac{E_1}{E_2} = \dfrac{\dfrac{1}{4\pi \epsilon_0} \dfrac{Q_1}{R_1^2}}{\dfrac{1}{4\pi \epsilon_0} \dfrac{Q_2}{R_2^2}} = \dfrac{Q_1}{Q_2} \times \left(\dfrac{R_2}{R_1}\right)^2$$

Given $$\dfrac{E_1}{E_2} = \dfrac{R_1}{R_2} \Rightarrow \dfrac{Q_1}{Q_2}\times \left(\dfrac{R_2}{R_1}\right)^2 = \dfrac{R_1}{R_2}$$

$$\Rightarrow \dfrac{Q_1}{Q_2} = \left(\dfrac{R_1}{R_2}\right)^3$$ ...(v)

$$\Rightarrow \dfrac{V_1}{V_2} = \dfrac{\dfrac{1}{4\pi \epsilon_0} \dfrac{Q_1}{R_1}}{\dfrac{1}{4\pi \epsilon_0} \dfrac{Q_2}{R_2}} = \dfrac{Q_1}{Q_2} \times \dfrac{R_2}{R_1} = \left(\dfrac{R_1}{R_2}\right)^3 \times \dfrac{R_2}{R_1}$$

$$\Rightarrow \dfrac{V_1}{V_2} = \left(\dfrac{R_1}{R_2}\right)^2$$
Question 5
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The electric field in a region of space is given by $$\vec E=E_0\hat i+2 E_o\hat j$$ where $$E_o=100 N/C$$. The flux of this field through a circular surface of radius $$0.02$$ m parallel to the Y-Z plane is nearly :

A
$$0.125 Nm^2/C$$

B
$$0.02 Nm^2/C$$

C
$$0.005 Nm^2/C$$

D
$$3.14 Nm^2/C$$

Solution

Answer is A.
We know that,
$$\phi= E.A Cos\theta$$ .
Since surface area is parallel to Y-Z plane therefore, y component of electric field will not contribute to flux through the circular surface.
So,
$$\phi= \pi r^{2} E_o = 3.14\times 0.0004\times 100= 0.125 Nm^{2}/C$$
Question 6
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Three charged particles $$A, B$$ and $$C$$ with charges $$-4q, 2q$$ and $$-2q$$ are present on the circumference of a circle of radius $$d$$. The charged particles $$A, C$$ and centre $$O$$ of the circle formed an equilateral triangle as shown in figure. Electric field at $$O$$ along x-direction is:

A
$$\dfrac{\sqrt{3}q}{4\pi \in_0d^2}$$

B
$$\dfrac{\sqrt{3}q}{\pi \in_0d^2}$$

C
$$\dfrac{2\sqrt{3}q}{\pi \in_0d^2}$$

D
$$\dfrac{3\sqrt{3}q}{4\pi \in_0d^2}$$

Solution

The electric field due to B will be in the direction of line joining B and O (or B and C) [Ref. image 1]

The electric field due to C will be in the direction of line joining O and C.

Net electric Field due to B and C along OC $$= \dfrac{k2q}{d^2} + \dfrac{k2q}{d^2} = \dfrac{k4q}{d^2}$$

Now, Electric field due to A along OA $$= \dfrac{k4q}{d^2}$$

Net Electric field along the x-axis is [Ref. image 2]

$$2 \left[\dfrac{k4q}{d^2}\cos 30^o\right] = \dfrac{k4\sqrt{3}q}{d^2}$$

$$= \dfrac{\sqrt{3}q}{\pi \in_0d^2}$$
Question 7
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Two identical conducting spheres $$A$$ and $$B$$, carry equal charge. They are separated by a distance much larger than their diameter, and the force between them is $$F$$. A third identical conducting sphere, $$C$$, is uncharged. Sphere $$C$$ is first touched to $$A$$, then to $$B$$, and the removed. As a result, the force between $$A$$ and $$B$$ would be equal to

A
$$\dfrac {3F}{4}$$

B
$$\dfrac {F}{2}$$

C
$$F$$

D
$$\dfrac {3F}{8}$$

Solution

Initially charges on both sphere, $$q$$

$$F = \dfrac{Kq^2}{d^2}$$ .............(1)

When sphere C will get touched with sphere A, then final charges on both will become $$\dfrac{q}{2}$$.

Now, when this sphere C will get touched to sphere B, then final charges on both of them will be
$$q_c = q_d = \dfrac{q/2 + q}{2} = \dfrac{3q}{4}$$

Now force between A and B will be
$$F' = \dfrac{k \times \dfrac{q}{2} \times \dfrac{3q}{4}}{d^2}$$
$$F' = \dfrac{3F}{8}$$
Question 8
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Consider an electric field $$\overline{E}=E_{0}\hat{x}$$, where $$E_{0}$$ is a constant. The flux through the shaded area (as shown in the figure) due to this field is:

A
$$2E_{0}a^{2}$$

B
$$\sqrt{2}E_{0}a^{2}$$

C
$$E_{0}a^{2}$$

D
$$\displaystyle \dfrac{E_{0}a^{2}}{\sqrt{2}}$$

Solution

$$Flux\ through\ area = Electric\ field \times Area\ projected$$
$$\phi =EA cos45^o$$
$$= E_0(\sqrt{2}a^2) \times \dfrac{1}{\sqrt2}$$
$$= E_0a^2$$
Question 9
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Two point charges A and B, having charges +Q and -Q respectively, are placed at certain distance apart and force acting between them is F. If 25% charge of A is transferred to B, then force between the charges becomes

A
$$F$$

B
$$\dfrac{9 F}{16}$$

C
$$\dfrac{16 F}{9}$$

D
$$\dfrac{4 F}{3}$$

Solution

$$F=\dfrac{KQ^2}{r^2}$$
If 25% of charge of A transferred to B then
$$q_A = Q - \dfrac{Q}{4} = \dfrac{3 Q}{4} $$ and $$q_B = -Q + \dfrac{Q}{4} = \dfrac{-3Q}{4}$$
$$F_1 = \dfrac{kq_A q_B}{r^2}$$
$$F_1 = \dfrac{k \left(\dfrac{3Q}{4} \right)^2}{r^2}$$
$$F_1 = \dfrac{9}{16} \dfrac{kQ}{r^2}$$
$$\implies F_1 = \dfrac{9F}{16}$$
Question 10
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Suppose the charge of a proton and an electron differ slightly. One of them is -e, the other is $$(e+\Delta e)$$. If the net of electrostatic force and gravitational force between two hydrogen atoms placed at a distance d (much greater than atomic size) apart is zero, then $$\Delta e$$ is of the order of [Given mass of hydrogen $$m_h=1.67\times 10^{-27}Kg$$]

A
$$10^{-47}C$$

B
$$10^{-20}C$$

C
$$10^{-23}C$$

D
$$10^{-37}C$$

Solution

Given : $$m_h= 1.67\times 10^{-27} \ kg$$
Net charge on hydrogen atom $$q = (e+\Delta e)-e = \Delta e$$
Let the distance between them be $$d$$.
Equating gravitational force and the electrostatic force, we get
$$\dfrac{Gm_h^2}{d^2} = \dfrac{k (\Delta e)^2}{d^2}$$
Or $$\dfrac{(6.67\times 10^{-11})(1.67\times 10^{-27})^2}{d^2} = \dfrac{(9\times 10^9)(\Delta e)^2}{d^2}$$
$$\implies \ \Delta e = 2.06\times 10^{-37} \ C$$
The correct answer is option D.