Electric Charges and Fields Test

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Electric Charges and Fields Test - 74

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Question 1
1 / -0

In Coulumb's law, the constant of proportionality, $$ K = \dfrac{1}{ 4 \pi \epsilon_0 } $$ has units

A
$$N$$

B
$$N-m^2 $$

C
$$Nm^2 / C ^2 $$

D
$$ NC^2 / m^2 $$

Solution

The Coulomb force is given as

$$F = \dfrac{{K{q_1}{q_2}}}{{{R^2}}}$$

The unit of charge is coulomb$$\left( C \right)$$

Unit of distance is meter $$\left( m \right)$$

Unit of Force is Newton $$\left( N \right)$$

Then on substituting all the

$$K = \dfrac{{N - {m^2}}}{{{C^2}}}$$
Question 2
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Two charges equal in magnitude and opposite in polarity are placed at a certain distance apart and force acting between them is $$F$$. If $$75$$% charge of one is transferred to another, then the force between the charges becomes

A
$$\cfrac{F}{16}$$

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

C
$$F$$

D
$$\cfrac{15F}{16}$$

Solution
Question 3
1 / -0

The force between two charges 0.06 m apart is $$5\ N$$. If each charge is moved towards the other by $$0.01\ m$$, then the force between them will become

A
$$7.20\ N$$

B
$$11.25\ N$$

C
$$22.50\ N$$

D
$$45.00\ N$$

Solution

$$\textbf{Step 1: Initial Force between two charges }$$
Using Coulomb's law, $$F = \dfrac {Kq_{1}q_{2}}{r^{2}}$$
$$\Rightarrow 5 = \dfrac {Kq_{1}q_{2}}{(0.06)^{2}} $$ $$.... (1)$$

$$\textbf{Step 2: New Force between the charges}$$
When each charge is moved towards the other by $$0.01m$$ , the new distance between them is $$r'= 0.06m-2\times 0.01m= 0.04m$$
$$F' = \dfrac {Kq_{1}q_{2}}{(0.04)^{2}}$$ $$.....(2)$$

$$\textbf{Step 3: Solving Equations}$$
Dividing $$(2)$$ by $$(1)\Rightarrow$$ $$\dfrac{F'}{5} = \dfrac {(0.06)^2}{(0.04)^{2}}$$
$$\Rightarrow F' = 11.25\ N$$

The new force would be $$11.25\ N$$
Hence, Option(B) is correct.
Question 4
1 / -0

A square surface of side $$L$$ metre in the plane of the paper is placed in a uniform electric field $$E(V/m)$$ acting along the same place at an angle $$\theta$$ with the horizontal side of the square as shown in figure. The electric flux linked to the surface in unit of $$V-m$$, is

A
$$EL^{2}$$

B
$$EL^{2}\cos \theta$$

C
$$EL^{2}\sin \theta$$

D
Zero

Solution
Question 5
1 / -0

Directions For Questions

A $$3\mu F$$ capacitor up to $$300\ volt$$ and $$2\mu F$$ is charged up to $$200\ volt$$. The capacitor are connected so that the plates of same polarity are connected together.

...view full instructions

if instead if this, the plates of this, the plates of opposite polarity were joined together, then amount of charge that flows is:

A
$$6 \times 10^{-4}\ C$$

B
$$1.5\times 10^{-4}\ C$$

C
$$3\times 10^{-4}\ C$$

D
$$7.5\times 10^{-4}\ C$$

Solution
Question 6
1 / -0

Two point charges of $$20 \mu C$$ and $$80 \mu C$$ are $$10cm$$ apart. Where will the electric field intensity be zero on the line joining the charges from $$20\mu C$$ charge? (in m)

A
$$0.04$$

B
$$0.1$$

C
$$0.033$$

D
$$0.33$$

Solution
Question 7
1 / -0

Two fixed charges A and B of $$5\mu C$$ is separated by a distance of 6 metre. C is the midpoint of the line joining A and B. A charge 'Q' of $$-5\mu C$$ is shot perpendicular to the line joining A and B through C with a kinetic energy of 0.06 J The charge Q comes at rest at a point D. The distance CD is

A
3m

B
13m

C
33m

D
4m

Solution
Question 8
1 / -0

A spherical shell of radius $$R$$ has two holes $$A$$ and $$B$$ through which a ring having charge positive through without touching the shell. The flux through spherical shell is $$\dfrac{q}{3 \varepsilon_0}$$. Find the radius of the ring:-

A
$$R\sqrt{2}$$

B
$$2 R$$

C
$$0.5 R$$

D
$$R$$

Solution
Question 9
1 / -0

Equal charges of each $$2\mu C$$ are placed at a point $$x=0,2,4,$$ and $$8$$ $$cm$$ on the $$x-$$axis. The force experienced by the charge at $$x=2\ cm$$ is equal

A
$$5$$ Newton

B
$$10$$ Newton

C
$$0$$ Newton

D
$$15$$ Newton

Solution
Question 10
1 / -0

N fundamental charges each of the charge '$$q$$' are to be distributed as two point charges separated by a fixed distance, then the maximum to minimum force bears a ratio ( N is even and greater than $$2$$)

A
$$\dfrac{{{{(N - 1)}^2}}}{{4{N^2}}}$$

B
$$\dfrac{{4{N^2}}}{{(N - 1)}}$$

C
$$\dfrac{{{N^2}}}{{4(N - 1)}}$$

D
$$\dfrac{{2{N^2}}}{{(N - 1)}}$$
Solution
Hint: Arithmetic mean is always greater or equal to Geometric mean.
Correct option: Option (C)
Explanation of correct option:

Step 1: Write the values of both charges.

As $$N$$ no. of charges are separated each with charge $$q$$, let $$n$$ be the no. of charges at the first point with $$q_1=nq$$ charge and thus another point will have $$(N-n)$$ no. of charges with charge $$q_2=(N-n)q$$.

Step 2: Apply Coulomb's law

Electrostatic force$$(F)$$ is directly proportion to the product of charges$$(q_1q_2)$$ and inversely proportional to square of distance between them$$(r^2)$$
$$F=k\dfrac{q_1q_2}{r^2}$$ ,,,$$(k= proportionality\ constant)$$

$$\therefore F=\dfrac{k(nq)(N-n)q}{r^2}=\dfrac{kq^2n(N-n)}{r^2}$$

Step 3:

As the no. of charges are natural numbers, the arithmetic mean is greater than equal to the geometric mean.
$$\therefore AM \ge GM$$

Arithmetic mean is the ratio of the sum of the values to the no. of values.
$$AM=\dfrac{n+(N-n)}{2}=\dfrac{N}{2}$$

Geometric mean is the $$n^{th}$$ root of the product of the $$n$$ no. of values.
$$GM=\sqrt{n(N-n)}$$

$$\therefore \dfrac{N}{2} \ge \sqrt{n(N-n)}$$

$$\therefore \dfrac{N^2}{4}\ge n(N-n)$$

Maximum value of $$n(N-n)=\dfrac{N^2}{4}$$

$$\therefore n(N-n)=\dfrac{N^2}{4}$$

$$Nn-n^2=\dfrac{N^2}{4}$$

$$n^2-Nn+\dfrac{N^2}{4}=0$$

$$n^2-Nn=\dfrac{N^2}{4}=(n-\dfrac{N}{2})^2=0$$

$$\therefore n=\dfrac{N}{2}$$

Step 4:

Maximum value of Force $$F_{max}=\dfrac{kq^2(\dfrac{N}{2})(N-\dfrac{N}{2})}{r^2}=\dfrac{kq^2N^2}{4r^2}$$

The minimum value of Force (n=1) $$F_{min}=\dfrac{kq^2(1)(N-1)}{r^2}=\dfrac{kq^2(N-1)}{r^2}$$

Taking the ratio of both the calculated forces,

$$\dfrac{F_{max}}{F_{min}}=\dfrac{\dfrac{kq^2N^2}{4r^2}}{\dfrac{kq^2(N-1)}{r^2}}=\dfrac{N^2r^2}{(N-1)4r^2}=\dfrac{N^2}{4(N-1)}$$

Thus, the ratio of maximum to minimum force is $$\dfrac{N^2}{4(N-1)}$$.